Let theta (which I'll denote by t here for brevity) be half of the smaller angle of the rhombus, at the vertex away from the square. Then, x/2 = sin(t) and y/2 = cos(t). The area of the rhombus is 2sin(t)cos(t) and l = sqrt(2)sin(t), so that the area of the square is l^2 = 2sin^2(t). Thus, the area of shaded pink region is 2sin(t)cos(t) - 2sin^2(t), which we will be maximize through clever use of the double angle formulas. 2sin(t)cost(t) - 2sin^2(t) = sin(2t) + cos(2t) - 1 by the sine and cosine double angle formulas. (sin(2t) + cos(2t)^2 = sin^2(2t) + 2sin(2t)cos(2t) + cos^2(2t) = 1 + sin(4t), so sin(2t) + cos(2t) - 1 = sqrt(1 + sin(4t)) - 1 (when t is between 0 and pi/4, like it is in our problem). sqrt(1 + sin(4t)) - 1 is clearly maximized when sin(4t) = 1, i.e. when 4t = pi/2 or t = pi/8. Then, l^2 = 2sin^2(t) = 1 - cos(2t) = 1 - cos(pi/4) = 1 - sqrt(2)/2, and l = sqrt(1 - sqrt(2)/2), just as you got. Edit: I felt all clever doing this on my own, just referencing the beginning of your video to springboard off your definitions of x/2 and y/2, only to see that at the end you basically told us to do it this way. :(
If you use the theta angle and vectors then you get: Area of the rhomboid:sin(theta) The opposite diagonal: (2-2cos(theta))^(1/2) Area of the square: 1-cos(theta) Then the area of the part that needs to be maximized is sin(theta)+cos(theta)-1 which is maximized when theta is pi/4, from there L can be calculated.
Without calculus, denoting theta as t here: Following, from the one angle in the triangle being 135° and the opposite site being equal to 1, the area of a fourth of the shaded area is A = sin(t/2)sin(45°-t/2) / 2sin(135°) (Derived from sine rule A=ab x sin(135°) and the three angles being t/2, 135° and 45°-t/2) ) to maximize A we need to maximize the product sin(t/2) * sin(45°-t/2) By symmetrie t/2 must be between 0 and 22,5° , but looking at slope of sin(x) clearly sin(t/2) is both smaller than sin(45°-t/2) and growing faster than sin(45°-t/2) is decreasing. Thus t/2=22,5° must hold to maximize the product. (Is this without calculus, when I am argumenting with slopes? Maybe not, I am afraid, still found this a neat solution.)
Let t be the left angle of the square, leaving the red area to sint-l² By the cosine law, we know cost=(1²+1²-2l²)/2×1×1=1-l² Which means, sint-l²=sint+cost-1 Since sin²t+cos²t=1, by Cauchy-Schwarz inequality, we can find (sint+cost)²
2:53 Not all parallelograms have diagonals that intersect at right angles. For instance, rectangles are a counter example. Rhombi, conveniently, do have this property.
Well, this is really easy to do with regular geometry. Call the rhombus ABCD and the square BEDF (same orientation). Note that A, F, E, C lie in this order on one line - the perpendicular bisector of BD. We notice that since AFB, BEC, CEF and DFA are all congruent, the area in question is just 4x the area of AFD. There are two key properties of this triangle. First is that one of its sides - AD, has height 1. The other is that the angle AFD=180-DFE=180-45=135. So we are trying to maximise the area of a triangle that has a side 1 and the opposite angle 135 degrees. It is known that for fixed AD, all possible points F lie on an arc of a circle from A to D - this is a consequence of an inscribed angle theorem. And obviously the point F maximizing the area according to an old formula base*height/2 for a fixed base AD=1 has to be the furthest possible distance from AD on this arc i.e. in the midpoint of this arc. Thus for AFD maximising the area we have to have AF=FD and since AFD=135 degrees ADF=DAF=22.5 degrees. So by these congruencies we get that the angles of the rhombus must be 45 and 135. To answer the question about l one just have to figure out the arm lengths for the 22.5, 22.5, 135 triangle with base 1 which is just simple trigonometry.
EDIT: to solve without calculus, work the geometry to get (A+L^2)^2 = 2L^2(1-L^2/2). Graphically both sides are parabolas in L^2. The LHS is upward curving with zero at A=-L^2, the RHS is downward curving with zeros at L^2=0, L^2=2. Imagine translating the LHS parabola along the L^2-axis. Maximum A will occur when the parabolas just touch. This occurs when the discriminator of the above quadratic, i.e. a^2 + 2A -1, is zero. This is a quadratic in A, solve it for A = 2^(1/2-1) and then the first one for L^2. Both roots are equal (because of the vanishing discriminator) and are given by L^2 = 1-1/2^(1/2)
Here is an arguably no-calculus, no-algebra solution: Let's label some points: A is the upper left corner of the rhombus B is the upper left corner of the square C is the upper right corner of the rhombus and square AC is 1. BC is l, the value to be determined Drop a perpendicular from B to AC, and call the intersection D. Consider triangle ABC. Its area is 1/4 of the purple area to be maximized, so we want to maximize the area of this triangle. If you consider AC as the base of the triangle and BD as its height, maximizing the area of the triangle (AC*BC/2) means maximizing BD (since AC is invariant 1). Some simple geometry will reveal that the angle ABC is 135 degrees no matter what l is. This means that points A, B, and C will always lie on a circle of a particular radius which is invariant (because AC and angle ABC are invariant). This is where there is perhaps a bit of hand-waving: The length of BD is maximized when D bisects the chord AC and B bisects the arc ABC. There, end of hand-waving. Did I have to use calculus to do this or is it obvious? Perhaps there is some geometric proof of this. At this point triangle ABC is isosceles, so angle BCA is half of (180-angle ABC), or 22.5 degrees, and the distance CD is 1/2 because D bisects AC. Simple trigonometry on right triangle BDC yields l = 1/(2*cos(22.5)) which you can evaluate using the half-angle formula for cos(45/2) knowing sin(45) = cos(45) = 1/sqrt(2) and the result is sqrt(1-1/sqrt(2)). (this was corrected from its original version where the height of the triangle was incorrectly referred to as being BC)
it's not hand-waving. First you get restriction: ABC is always 135 degrees. Then, you draw new picture: AC - base (horizontally) - fixed length = 1. And B lie on fixed circle. Then, you need to maximize height of triangle ABC dropped from point B. Higher point B -> higher height. So we need to pick 'most upwards' point on the circle. And this is opposite point to center of segment, so ABC is isosceles.
Denote by x half the small diagonal and y half the large diagonal of the rhombus. Note that x is also half the diagonal of the square. By Pythagoras theorem x^2 + y^2 = 1 and we need to maximize 2xy - 2x^2. Let k = 1 + sqrt(2) be the positive root of the equation X^2 - 2X - 1 = 0. By AM - GM 2xy - 2x^2 = 1/k * 2 * kx * y - 2x^2
It made me smile when you used Pythag to derive the area of the square in terms of x, having started by giving the formula for the area of a rhombus, of which a square is just a special case.
θ = (angle between one side of a square and one side of a rhombus) x/2 = cos(θ+π/4) y/2-x/2 = sqrt(2)sin(θ) A = 4*1/2*x/2*(y/2-x/2) = 2sqrt(2)*sin(θ)cos(θ+π/4) = 2sqrt(2)*[sin(θ+(θ+π/4))+sin(θ-(θ+π/4))] = 2sqrt(2)*[sin(2θ+π/4)-1/sqrt(2)] is maximal: 2θ+π/4=π/2, θ=π/8 therfore, x = 2cos(3π/8) = 2sqrt(1/2-1/2sqrt(2)) = sqrt(2-sqrt(2))
1. Define the angle on rhombus next to the square's right angle to be theta 2. By trigonometry and the formula A=1/2absin(C), you'll get Area=2sin(theta)(cos(theta)-sin(theta)) 3. By using double angle formula Area=sin(2theta)+cos(2theta)-1=sqrt(2)sin(2theta+pi/4)-1 4. sin(pi/2)=1 is max so 2theta+pi/4=pi/2 =>theta = pi/8=22.5 deg 5. Therefore, Area=sqrt(2)-1 and square side length = sqrt(1-1/sqrt(2))
If the angle of rhombus is 2α, then l=√2sin(α), area of rhombus is sin(2α), area of square is 2sin²α. A=sin(2α)-2sin²α=sin(2α)+cos(2α)-1=√2cos(2α-π/4)-1≤√2-1. A=√2-1 for α=π/8, and l=√2sin(π/8)=√(1-cos(π/4))=√(1-1/√2).
To solve it without calculus, you need to notice that you're maximizing area of triangle between upper left corner of the rhombus, upper left corner of the square, and upper right corner of the square. Now its sides are l and 1 and the angle between horizontal side and side going to upper left corner of the rhombus is 135 degrees. And we're kind of sliding the side of length 1 between these two lines, so obviously maximum is where they make an isosceles triangle (maximizes height with given base of length 1), therefore the two remaining angles are 22.5 degrees. And that gives us l = 1/(2 cos(22.5)) with the angle in degrees. Edit: oh, and the angle theta is obviously 45 degrees, twice the 22.5.
If the apex angle of the acute side of the rhombus is θ, then from the cosine rile cosθ=1-L^2 Red area: S=sinθ-L^2=sinθ+cosθ-1 ∴dS/dθ=sinθ-cosθ=√2sin( θ+135°) ∴S is maximum at θ=45° L=√(1-1/√2)
We simply take x = 2*sin(t). Then A = - 2*sin^2(t) + 2*sin(t)*cos(t) = - 1 + cos(2*t) + sin(2*t) = - 1 + sqrt(2) *sin(2*t + pi/4). To maximize A, we have to maximize the sine function. Thus, 2*t + pi/4 = pi/2 => t = pi/8 and Amax = - 1 + sqrt(2). Then, x = 2*sin(pi/8) = sqrt(2- sqrt(2)) and l = x/sqrt(2) = sqrt(1-1/sqrt(2)).
There are 4 equal purple triangles, so it's enough to max the area of one. Area of such triangle = 1/2 *base*(L/sqrt2); meaning that value of (base * L) should be maximized. Thus: base = L. (L/sqrt2)^2 + (L + L/sqrt2)^2 = 1; and L is solvable.
Using the substitution x = 2 sin u (u ∊ (0, ¼π)) (u is half the angle Michael was talking about btw), we get 🟪 = 2 sin u cos u - 2 sin^2 u = sin 2u + cos 2u - 1 = √2 cos(2u - ¼π) - 1 Notice that the argument of cosine ranges from -¼π to +¼π, but we look at the graph of cosine (or at a circle), cos is max when the argument is 0 Hence, max(🟪) = √2 - 1 θ = 2u = ¼π
divide the region into 4 triangles with sides l, 1, and k. side k is along one of the diagonals. now since the diagonals of the rhombus and square are colinear, then it divides the interior angle of the square into two 45 degrees. this means the angle between l and k is 135 degrees. now, to max the area of the region, we need to max the area of the triangle. its area is given by kl sin 45. so to mx the area, we need to max the product of k and l using law of cosine, we see that 1=k^2+l^2+kll sqrt(2). by completing the square and rearranging, we see that 1-(k-l)^2=kl times some constant. so to max kl, we need to set k-l to zero. this means k=l. so the triangle is an isosceles triangle with angles 135, 22.5 and 22.5 degrees. using half angle identity we can find the value of l.
I figured it out in terms of angle. Area of rhomb is sin(a), short diagonal is 2*sin(a/2), area square of square is 2*(sin(a/2))^2. Shaded area is sin(a)-2*(sin(a/2))^2=sin(a)-1+cos(a), deriv. is cos(a)-sin(a). a=pi/4
Area of the square is 1/2*x*x ... because the square is just a special case of a rhombus and the area (as noted) is just 1/2 the product of the diagonals.
alternative solution, uses trig, but as far as I know, no calculus: let x be the smaller angle of the rhombus. the area of the rhombus is sin x = base x perp. height. also, by cosine law, we have 2 l^2 = 2-2cos x. so the area of the square = l^2 = 1-cos x. the remaining area is sin x + cos x -1, the last constant can be discarded for the maximization. maximizing the first two is the same as maximizing their square sin^2 x + 2 sin x cos x + cos^2 x = 1 + sin(2x). this is maximized when the sine term = 1, which is at x = pi/4 = 45 degrees. since l^2 = 1 - cos pi/4, we have l = sqrt( 1 - cos (pi/4)) = sqrt( 1 - 1/sqrt(2)).
With the geometry we can find that the area is L√(2-L^2) - L^2 Let L=√(2)sin(x), the area will become 2sin(x)cos(x) - 2sin(x)^2 = sin(2x) + cos(2x) - 1 = √(2)sin(2x + π/4) - 1 Because 0 < L < 1, we have 0 < x < π/4 Therefore the maximal area is √(2)sin(π/4 + π/4) - 1 = √(2) - 1 when L = √(2)sin(π/8)
Let the acute angle of rombus be 2x Area of Rombus = 2 sinx cosx Area of square = 2 sinx^2 Shared area = 2sinx cosx - 2 sinx^2 = sin2x + cos2x - 1 = sin2x + sin (pi/2-2x) - 1 = 2sin(pi/4)cos(2x-pi/4) - 1 Maximum happens when 2x = pi/4 (45 degrees) Now you just need to find L = sqrt(2) sin pi/8, which you can do with cos 2x = 1 - 2 sinx^2
Well knowing that square is rhombus and that both diagonals are x it is easy to see that area is (x^2)/2 using the rhombus formula... Finding relationship between x and l is required only for final answer really
There's another way to find the maximum of the function without calculating the derivative: If we call x = 2 sin z, with z in [0, pi/4], we get the function A(x(z)) = 2(sin z * cos z - sin^2 z) Once 2sin^2 z = 1-cos 2z, and 2sin z cos z = sin 2z A(z) = -1 + sin 2z + cos 2z = -1 + sqrt(2) * sin (2z + pi/4) We get the maximum with sin is the maximum: 2z + pi/4 = pi/2 z = pi/8 x = 2 * sin (pi/8) L = sqrt(2) * sin(pi/8)
Area of rhombus equals base x height = sin(theta). By cosine rule X^2 = 2 - 2 * cos(theta). Area of square = 0.5 * X^2 = 1 - cos(theta). Area of square is also L^2 so L = sqrt(1-cos(theta)). Wanted area = sin(theta) + cos(theta) - 1; derivative cos(theta) - sin(theta) = 0; tan(theta) = 1; theta = pi / 4, cos(theta) = 1/ sqrt(2). So L = sqrt(1-1/sqrt(2)).
That's how I did it too, and you don't need calculus to show that there is a maximum in sin(t) + cos(t) - 1 at pi/4. A sketch or even intuition is enough, but formally: sin(t) + cos(t) = sqrt[(sin(t)+cos(t)^2)] = sqrt(sin^2(t) + cos^2(t) + 2sin(t)cos(t)) = sqrt(1 + 2sin(t)cos(t)) = sqrt(1 + sin(2t)), which has a maximum when 2t = pi/2. We need to be careful about using x = sqrt(x^2) if x might be negative, of course, but here we can see that it won't be in the domain 0
My guess for this maximum area problem geometrically is when the rhombus is a square or when one of the angles is a convenient 45°. Generally it will need an ingenious technique. That's how it usually works out in such problems.
So, it's easier without calculus actually. Having the eq. 2A = 2*sqrt{4-x²}-x², you take A as a constant, and you obtain a quadratic equation for x². For maximizing it you impose the discriminant is zero, so A= sqrt{2}-1. Then x² = 1- A, which is your result, and then replace its value in l, and terminado.
First the problem to solve these algebraically and/or geometrically is generally not easy! Second, that is the power of the Calculus! So so so much easier with it! Recall Kepler plodded for decades before he hit on his laws. If he had made any numerical or other mistakes he probably wouldn't have gotten his monumental results! Imagine if he had the Calculus a century or so later he would have avoided that drudery!
I took the challenge, solved with linear algebra. Haven't watched your approach yet. If I find that we solved it different ways, I will type up my solution since this was a really fun problem, at lease the way I took it.
Without calculus, denoting theta as t here: Following, from the one angle in the triangle being 135° and the opposite site being equal to 1, the area of a fourth of the shaded area is A = sin(t/2)sin(45°-t/2) / 2sin(135°) (Derived from sine rule A=ab x sin(135°) and the three angles being t/2, 135° and 45°-t/2) ) to maximize A we need to maximize the product sin(t/2) * sin(45°-t/2) By symmetrie t/2 must be between 0 and 22,5° , but looking at slope of sin(x) clearly sin(t/2) is both smaller than sin(45°-t/2) and growing faster than sin(45°-t/2) is decreasing. Thus t/2=22,5° must hold to maximize the product. (Is this without calculus, when I am argumenting with slopes? Maybe not, I am afraid, still found this a neat solution.)
Let’s call the points of the rhombus ABCD, and the other two points of the square EF (so now it’s BEDF) let M be the center of AC, and theta be the angle MAB. a^2+(atan(theta))^2=1 (tan(theta)+1)a^2=1 a^2=1/(tan(theta)+1) a^2*tan(theta)=tan(theta)/(tan(theta)+1) a=1/sqrt(tan(theta)+1) atan(theta)=tan(theta)/sqrt(tan(theta)+1) Area of ABCD=2tan(theta)/(tan(theta)+1) sqrt(2)l=2tan(theta)/sqrt(tan(theta)+1) l=sqrt(2)tan(theta)/sqrt(tan(theta)+1) l^2=2tan(theta)^2/(tan(theta)+1) we want to maximize (2tan(theta)/tan(theta)+1)-2tan(theta)^2/(tan(theta)+1) If tan(theta)+1>1, we can multiply by tan(theta)+1 and maximize 2tan(theta)-tan(theta)^2 We can let t be tan(theta) We want to maximize 2t-t^2=2t(t-1) This is a parabola with zeroes at 0 and 1 so its maximum is at t=1/2, which makes l be sqrt(2)(1/2)/sqrt(1/2+1)=1/2sqrt(3)
I also have a challenge: 11:10 Represent √( 2 - √2 ) as (a - b*√2), where a>0, b>0. Is it even possible? If NO, explain, why. If YES, calculate "a" and "b".
It's possible. (a-bsqrt2)²=(a²+b²)-2absqrt2=2-sqrt2 a²+b²=2 ab=1/2 Therefore, a²+b²+2ab=(a+b)²=3=(a+b)² Since a>0, b>0, a+b=sqrt3 If we see a,b as solutions of quadratic equation (x-a)(x-b)=0, we can say 2x²-2sqrt3×x+1=0 Then we can conclude x=(sqrt3-1)/2 or x=(sqrt3+1)/2 if a
@@김강민-j8x Thank yoy! Your appoach looks still easier than mine. But you made an important mistake: The first equation is NOT "a²+b²=2" but it should be "a²+2*b²=2" (as we square not "b" but "b*sqrt(2)"). I always did it like that: 1) a² + 2b² = 2 2) 2ab = 1 And I solved this system of equations From the second equation, we have 2) b = 1 / (2a) And we substitute this into the first equation: 1) a² + 2b² = 2 a² + 2 * [1 / (2a)]² = 2 a² + 2 * [1 / (4a²)] = 2 || * 4a² 4a^4 + 2 = 8a² 4a^4 - 8a² + 2 = 0 || :2 2a^4 - 4a² + 1 = 0 2*(a²)² - 4a² + 1 = 0 Substitute t=a² (t>=0): 2*(t)² - 4t + 1 = 0 And solving this quadratic equation for "t" we get: Discriminant = DELTA = (-4)² - 4*2*1 = 16 - 8 = 8 sqrt(DELTA) = sqrt(8) = 2*sqrt(2) t.1 = [4 - 2*sqrt(2)] / 4 = [2 - sqrt(2)] / 2 Since 2 > sqrt(2), there is [2 - sqrt(2)] > 0 and t.1 > 0, which is O.K. or t.2 = [4 + 2*sqrt(2)] / 4 = [2 + sqrt(2)] / 2 Since [2 + sqrt(2)] > 0, there is t.2 > 0, which is O.K. So we get two cases: case 1: a² = t.1 a² = [2 - sqrt(2)] / 2 Since a>0: a = sqrt{ [2 - sqrt(2)] / 2 } Which is complicated... :( Now I get another problem HOW TO REPRESENT THE NUMBER "sqrt{ [2 - sqrt(2)] / 2 }" in the form of "c - d*sqrt(2)", where c,d>0? And the second case: case 2: a² = t.2 a² = [2 + sqrt(2)] / 2 Since a>0: a = sqrt{ [2 + sqrt(2)] / 2 } Which is complicated... :( Now I get another problem HOW TO REPRESENT THE NUMBER "sqrt{ [2 + sqrt(2)] / 2 }" in the form of "e + f*sqrt(2)", where e,f>0?
intuition tells me instantly this area is maxed out when the rhombus is made of two equilateral triangles ie: diagonal of the square is length 1 => L = sqrt(2)/2 now let's watch the video :D
Ya I started by noticing that this is like how the angle to a chord from the center is twice the angle from the opposite point. So I knew the acute angle of the rhombus was 45 degrees. Then it was simple.
7:25; when I had used the chain rule with that equation, it left me with -2x/2(4-x^2)^1/2, not -x^2/2(4-x^2)^1/2, so I believe that your answer is wrong.
I am studying geometrical drawing, this challenge is perfect for my study, thanks YT algorithm, and thanks Michael Penn for making me remember why I love calculus
My solution didn't use calculus and is *very* informal, so it's not a method I would recommend to others. Consider the acute angle of the rhombus. If that angle is 0 then the area we're looking to maximize is 0, as that's the area of the whole rhombus. If that angle is 90, then that area is also 0, as now it's just overlapping the square. Obviously we want something in the middle, therefore the angle is 45 degrees. From there, use the law of cosines to find the diagonal of the square: d^2=2-sqrt(2), which makes l=sqrt[2-sqrt(2)]/sqrt(2)
Using θ, A = 4sin(θ/2)cos(θ/2) - 2sin(θ/2)²... crud, I gotta remember those double angle trig functions, because I'm pretty sure this reduces nicely for λ = θ/2.
"a rhombus is a parallellogram with all equal sides" is a partially redundant definition. "a rhombus is a quadrilateral with all equal sides" is better. You don't need parallellograms to define rhombi.
Well at 2.53 u said that diagonals of parallelograms intersect each other at right angles which is actually wrong since only the diagonals of square and rhombus intersect each other at right triangles but not that of others like rectangle and a simple parallelogram.
Solved it without calculus... but I get the wrong answer! I tried it by using the Arithmetic Geometric Inequality Theorem. The idea is, the max area occurs when the arithmetic and geometric means are equal, right? Let A = L and B = sqrt(2-L^2) - L Then you're trying to get (A+B)/2 = sqrt(AB), where AB = the area of the shaded part. The answer works out nicely, is very easy to calculate... and is WRONG. Why? EDIT: I think I understand my issue. Setting them equal will minimize the arithmetic mean (what I don't want), which is different from maximizing the geometric mean (what I do want). But I don't know how to fix it.
Hi I was wondering if you could show a solution to a problem of finding the radius of the small circle that is formed in the area between 3 larger but different touching circles please? Thus radius of small circle, r (the unknown) in terms of the large touching circles, R1, R2, & R3?
I solved it with Trigonometry without Calculus. At the end it will need to find maximum value for sin (x) + cos (x) . It is my common knowledge the x will be 45 degree . but to prove , this is the way . 1/square root of 2 = 1/1.414 = sin 45 = cos 45 . Re-write equation by 1.414 * [cos 45 * sin x + sin 45 * cos x ] = sin (x) + cos (x) . then pull out this formula sin (x+y) = sin x cos y + cos x sin y . 1.414 * [cos 45 * sin x + sin 45 * cos x ] = 1.414 * sin(45 + x) . Maximum sin = 1 . then x =45 degree. Math is funny . I got theta first then work my way to get " L" . Micheal get " L" first without need to find theta.
Yes, but you can solve this problem without derivatives and therefore without calculus. It comes to a curve touching problem, which I showed a little way down the thread. It's a very insightful comment (a bit like Merry at the Moria-gate), as the "curve-touching" approach leads in to a whole wonderland of algebraic curve theory, which is pursued in the spirit of projective geometry, which eschews >>limits
You are making a big mistake of setting a square inside a rumbas. If you put the opposite vertices of the square on the opposite vertices of the rumbas, the other two vertices of the square are alternatively locating on the long diagonal of the rumbas. That is because the long diagonal of the rumbas is a perpendicular bisector of its smaller diagonal. Unless you are thinking of a Parallelogram and not a rumbas. So no matter how you draw your square, its other two vertices must fall on the long diagonal of the rumbas. Dr. Zomorrodian, Mathematics professor
You said solve "without calculus", then you used derivatives. Okay, I guess that's technically pre-Calc, but I assumed "without calc" meant without derivatives because who on Earth would have though to use integrals on this???
-> Challenge: Solve this without Calculus! -> We solve a nice geometry problem using methods from first semester calculus. That was a good place to stop this bs.
My challenge to you! Solve without calclulus.
45°
Let theta (which I'll denote by t here for brevity) be half of the smaller angle of the rhombus, at the vertex away from the square. Then, x/2 = sin(t) and y/2 = cos(t). The area of the rhombus is 2sin(t)cos(t) and l = sqrt(2)sin(t), so that the area of the square is l^2 = 2sin^2(t). Thus, the area of shaded pink region is 2sin(t)cos(t) - 2sin^2(t), which we will be maximize through clever use of the double angle formulas.
2sin(t)cost(t) - 2sin^2(t) = sin(2t) + cos(2t) - 1 by the sine and cosine double angle formulas. (sin(2t) + cos(2t)^2 = sin^2(2t) + 2sin(2t)cos(2t) + cos^2(2t) = 1 + sin(4t), so sin(2t) + cos(2t) - 1 = sqrt(1 + sin(4t)) - 1 (when t is between 0 and pi/4, like it is in our problem). sqrt(1 + sin(4t)) - 1 is clearly maximized when sin(4t) = 1, i.e. when 4t = pi/2 or t = pi/8. Then, l^2 = 2sin^2(t) = 1 - cos(2t) = 1 - cos(pi/4) = 1 - sqrt(2)/2, and l = sqrt(1 - sqrt(2)/2), just as you got.
Edit: I felt all clever doing this on my own, just referencing the beginning of your video to springboard off your definitions of x/2 and y/2, only to see that at the end you basically told us to do it this way. :(
If you use the theta angle and vectors then you get:
Area of the rhomboid:sin(theta)
The opposite diagonal: (2-2cos(theta))^(1/2)
Area of the square:
1-cos(theta)
Then the area of the part that needs to be maximized is sin(theta)+cos(theta)-1 which is maximized when theta is pi/4, from there L can be calculated.
Without calculus, denoting theta as t here:
Following, from the one angle in the triangle being 135° and the opposite site being equal to 1, the area of a fourth of the shaded area is
A = sin(t/2)sin(45°-t/2) / 2sin(135°) (Derived from sine rule A=ab x sin(135°) and the three angles being t/2, 135° and 45°-t/2) )
to maximize A we need to maximize the product sin(t/2) * sin(45°-t/2)
By symmetrie t/2 must be between 0 and 22,5° , but looking at slope of sin(x) clearly sin(t/2) is both smaller than sin(45°-t/2) and growing faster than sin(45°-t/2) is decreasing. Thus t/2=22,5° must hold to maximize the product.
(Is this without calculus, when I am argumenting with slopes? Maybe not, I am afraid, still found this a neat solution.)
Let t be the left angle of the square, leaving the red area to sint-l²
By the cosine law, we know cost=(1²+1²-2l²)/2×1×1=1-l²
Which means, sint-l²=sint+cost-1
Since sin²t+cos²t=1, by Cauchy-Schwarz inequality, we can find (sint+cost)²
2:53 Not all parallelograms have diagonals that intersect at right angles. For instance, rectangles are a counter example. Rhombi, conveniently, do have this property.
Well, this is really easy to do with regular geometry. Call the rhombus ABCD and the square BEDF (same orientation). Note that A, F, E, C lie in this order on one line - the perpendicular bisector of BD. We notice that since AFB, BEC, CEF and DFA are all congruent, the area in question is just 4x the area of AFD. There are two key properties of this triangle. First is that one of its sides - AD, has height 1. The other is that the angle AFD=180-DFE=180-45=135. So we are trying to maximise the area of a triangle that has a side 1 and the opposite angle 135 degrees. It is known that for fixed AD, all possible points F lie on an arc of a circle from A to D - this is a consequence of an inscribed angle theorem. And obviously the point F maximizing the area according to an old formula base*height/2 for a fixed base AD=1 has to be the furthest possible distance from AD on this arc i.e. in the midpoint of this arc. Thus for AFD maximising the area we have to have AF=FD and since AFD=135 degrees ADF=DAF=22.5 degrees. So by these congruencies we get that the angles of the rhombus must be 45 and 135. To answer the question about l one just have to figure out the arm lengths for the 22.5, 22.5, 135 triangle with base 1 which is just simple trigonometry.
Very elegant solution!
4:11 remembering that a square is a special case of a rhombus we can jump directly to Asquare = (x^2)/2 using the previous formula.
EDIT: to solve without calculus, work the geometry to get (A+L^2)^2 = 2L^2(1-L^2/2). Graphically both sides are parabolas in L^2. The LHS is upward curving with zero at A=-L^2, the RHS is downward curving with zeros at L^2=0, L^2=2. Imagine translating the LHS parabola along the L^2-axis. Maximum A will occur when the parabolas just touch. This occurs when the discriminator of the above quadratic, i.e. a^2 + 2A -1, is zero. This is a quadratic in A, solve it for A = 2^(1/2-1) and then the first one for L^2. Both roots are equal (because of the vanishing discriminator) and are given by L^2 = 1-1/2^(1/2)
2:53 Diagonals of rhombi are perpendicular, not necessarily parallelograms.
Here is an arguably no-calculus, no-algebra solution:
Let's label some points:
A is the upper left corner of the rhombus
B is the upper left corner of the square
C is the upper right corner of the rhombus and square
AC is 1.
BC is l, the value to be determined
Drop a perpendicular from B to AC, and call the intersection D.
Consider triangle ABC. Its area is 1/4 of the purple area to be maximized, so we want to maximize the area of this triangle.
If you consider AC as the base of the triangle and BD as its height, maximizing the area of the triangle (AC*BC/2) means maximizing BD (since AC is invariant 1).
Some simple geometry will reveal that the angle ABC is 135 degrees no matter what l is. This means that points A, B, and C will always lie on a circle of a particular radius which is invariant (because AC and angle ABC are invariant).
This is where there is perhaps a bit of hand-waving: The length of BD is maximized when D bisects the chord AC and B bisects the arc ABC. There, end of hand-waving. Did I have to use calculus to do this or is it obvious? Perhaps there is some geometric proof of this.
At this point triangle ABC is isosceles, so angle BCA is half of (180-angle ABC), or 22.5 degrees, and the distance CD is 1/2 because D bisects AC.
Simple trigonometry on right triangle BDC yields l = 1/(2*cos(22.5)) which you can evaluate using the half-angle formula for cos(45/2) knowing sin(45) = cos(45) = 1/sqrt(2) and the result is sqrt(1-1/sqrt(2)).
(this was corrected from its original version where the height of the triangle was incorrectly referred to as being BC)
it's not hand-waving. First you get restriction: ABC is always 135 degrees. Then, you draw new picture: AC - base (horizontally) - fixed length = 1. And B lie on fixed circle. Then, you need to maximize height of triangle ABC dropped from point B. Higher point B -> higher height. So we need to pick 'most upwards' point on the circle. And this is opposite point to center of segment, so ABC is isosceles.
This is the most elegant solution of this entire comment section!
@@bigbrain296 Nah, *hydra147147* explains it in a much better way further down
The height of the triangle should be called BD in your notation.
@@giacomolanza1726 Well spotted. It should indeed be BD and not BC. I'll correct my comment when I have a chance.
Denote by x half the small diagonal and y half the large diagonal of the rhombus. Note that x is also half the diagonal of the square. By Pythagoras theorem x^2 + y^2 = 1 and we need to maximize 2xy - 2x^2. Let k = 1 + sqrt(2) be the positive root of the equation X^2 - 2X - 1 = 0. By AM - GM
2xy - 2x^2 = 1/k * 2 * kx * y - 2x^2
How do you choose k
@@Noname-67 I guess equality occurs when y = kx and choose k such that k - 2 = 1/k.
and that's a good place
I agree
It made me smile when you used Pythag to derive the area of the square in terms of x, having started by giving the formula for the area of a rhombus, of which a square is just a special case.
θ = (angle between one side of a square and one side of a rhombus)
x/2 = cos(θ+π/4)
y/2-x/2 = sqrt(2)sin(θ)
A = 4*1/2*x/2*(y/2-x/2)
= 2sqrt(2)*sin(θ)cos(θ+π/4)
= 2sqrt(2)*[sin(θ+(θ+π/4))+sin(θ-(θ+π/4))]
= 2sqrt(2)*[sin(2θ+π/4)-1/sqrt(2)]
is maximal: 2θ+π/4=π/2, θ=π/8
therfore,
x = 2cos(3π/8)
= 2sqrt(1/2-1/2sqrt(2))
= sqrt(2-sqrt(2))
1. Define the angle on rhombus next to the square's right angle to be theta
2. By trigonometry and the formula A=1/2absin(C), you'll get Area=2sin(theta)(cos(theta)-sin(theta))
3. By using double angle formula Area=sin(2theta)+cos(2theta)-1=sqrt(2)sin(2theta+pi/4)-1
4. sin(pi/2)=1 is max so 2theta+pi/4=pi/2 =>theta = pi/8=22.5 deg
5. Therefore, Area=sqrt(2)-1 and square side length = sqrt(1-1/sqrt(2))
2:54 Diagonals of parallelograms do not necessarily intersect at right angles. If the parallelogram is a rhombus (as in this case), then they do.
If the angle of rhombus is 2α, then l=√2sin(α), area of rhombus is sin(2α), area of square is 2sin²α. A=sin(2α)-2sin²α=sin(2α)+cos(2α)-1=√2cos(2α-π/4)-1≤√2-1. A=√2-1 for α=π/8, and l=√2sin(π/8)=√(1-cos(π/4))=√(1-1/√2).
To solve it without calculus, you need to notice that you're maximizing area of triangle between upper left corner of the rhombus, upper left corner of the square, and upper right corner of the square. Now its sides are l and 1 and the angle between horizontal side and side going to upper left corner of the rhombus is 135 degrees. And we're kind of sliding the side of length 1 between these two lines, so obviously maximum is where they make an isosceles triangle (maximizes height with given base of length 1), therefore the two remaining angles are 22.5 degrees. And that gives us l = 1/(2 cos(22.5)) with the angle in degrees.
Edit: oh, and the angle theta is obviously 45 degrees, twice the 22.5.
If the apex angle of the acute side of the rhombus is θ, then from the cosine rile cosθ=1-L^2 Red area: S=sinθ-L^2=sinθ+cosθ-1 ∴dS/dθ=sinθ-cosθ=√2sin( θ+135°) ∴S is maximum at θ=45° L=√(1-1/√2)
We simply take x = 2*sin(t). Then A = - 2*sin^2(t) + 2*sin(t)*cos(t) = - 1 + cos(2*t) + sin(2*t) = - 1 + sqrt(2) *sin(2*t + pi/4). To maximize A, we have to maximize the sine function. Thus, 2*t + pi/4 = pi/2 => t = pi/8 and Amax = - 1 + sqrt(2). Then, x = 2*sin(pi/8) = sqrt(2- sqrt(2)) and l = x/sqrt(2) = sqrt(1-1/sqrt(2)).
There are 4 equal purple triangles, so it's enough to max the area of one.
Area of such triangle = 1/2 *base*(L/sqrt2); meaning that value of (base * L) should be maximized. Thus: base = L.
(L/sqrt2)^2 + (L + L/sqrt2)^2 = 1; and L is solvable.
Using the substitution x = 2 sin u (u ∊ (0, ¼π)) (u is half the angle Michael was talking about btw), we get
🟪 = 2 sin u cos u - 2 sin^2 u
= sin 2u + cos 2u - 1
= √2 cos(2u - ¼π) - 1
Notice that the argument of cosine ranges from -¼π to +¼π, but we look at the graph of cosine (or at a circle), cos is max when the argument is 0
Hence,
max(🟪) = √2 - 1
θ = 2u = ¼π
divide the region into 4 triangles with sides l, 1, and k. side k is along one of the diagonals. now since the diagonals of the rhombus and square are colinear, then it divides the interior angle of the square into two 45 degrees. this means the angle between l and k is 135 degrees. now, to max the area of the region, we need to max the area of the triangle. its area is given by kl sin 45. so to mx the area, we need to max the product of k and l
using law of cosine, we see that 1=k^2+l^2+kll sqrt(2). by completing the square and rearranging, we see that 1-(k-l)^2=kl times some constant. so to max kl, we need to set k-l to zero. this means k=l. so the triangle is an isosceles triangle with angles 135, 22.5 and 22.5 degrees.
using half angle identity we can find the value of l.
I figured it out in terms of angle. Area of rhomb is sin(a), short diagonal is 2*sin(a/2), area square of square is 2*(sin(a/2))^2. Shaded area is sin(a)-2*(sin(a/2))^2=sin(a)-1+cos(a), deriv. is cos(a)-sin(a). a=pi/4
Area of the square is 1/2*x*x ... because the square is just a special case of a rhombus and the area (as noted) is just 1/2 the product of the diagonals.
alternative solution, uses trig, but as far as I know, no calculus:
let x be the smaller angle of the rhombus. the area of the rhombus is sin x = base x perp. height. also, by cosine law, we have 2 l^2 = 2-2cos x. so the area of the square = l^2 = 1-cos x. the remaining area is sin x + cos x -1, the last constant can be discarded for the maximization.
maximizing the first two is the same as maximizing their square sin^2 x + 2 sin x cos x + cos^2 x = 1 + sin(2x). this is maximized when the sine term = 1, which is at x = pi/4 = 45 degrees. since l^2 = 1 - cos pi/4, we have l = sqrt( 1 - cos (pi/4)) = sqrt( 1 - 1/sqrt(2)).
With the geometry we can find that the area is L√(2-L^2) - L^2
Let L=√(2)sin(x), the area will become 2sin(x)cos(x) - 2sin(x)^2 = sin(2x) + cos(2x) - 1 = √(2)sin(2x + π/4) - 1
Because 0 < L < 1, we have 0 < x < π/4
Therefore the maximal area is √(2)sin(π/4 + π/4) - 1 = √(2) - 1 when L = √(2)sin(π/8)
Let the acute angle of rombus be 2x
Area of Rombus = 2 sinx cosx
Area of square = 2 sinx^2
Shared area = 2sinx cosx - 2 sinx^2 = sin2x + cos2x - 1 = sin2x + sin (pi/2-2x) - 1 = 2sin(pi/4)cos(2x-pi/4) - 1
Maximum happens when 2x = pi/4 (45 degrees)
Now you just need to find L = sqrt(2) sin pi/8, which you can do with cos 2x = 1 - 2 sinx^2
Where did the "without calculus" part come in?
It's a challenge for us viewers.
Well knowing that square is rhombus and that both diagonals are x it is easy to see that area is (x^2)/2 using the rhombus formula... Finding relationship between x and l is required only for final answer really
There's another way to find the maximum of the function without calculating the derivative:
If we call x = 2 sin z, with z in [0, pi/4], we get the function A(x(z)) = 2(sin z * cos z - sin^2 z)
Once 2sin^2 z = 1-cos 2z, and 2sin z cos z = sin 2z
A(z) = -1 + sin 2z + cos 2z = -1 + sqrt(2) * sin (2z + pi/4)
We get the maximum with sin is the maximum:
2z + pi/4 = pi/2
z = pi/8
x = 2 * sin (pi/8)
L = sqrt(2) * sin(pi/8)
Area of rhombus equals base x height = sin(theta). By cosine rule X^2 = 2 - 2 * cos(theta). Area of square = 0.5 * X^2 = 1 - cos(theta). Area of square is also L^2 so L = sqrt(1-cos(theta)). Wanted area = sin(theta) + cos(theta) - 1; derivative cos(theta) - sin(theta) = 0; tan(theta) = 1; theta = pi / 4, cos(theta) = 1/ sqrt(2). So L = sqrt(1-1/sqrt(2)).
That's how I did it too, and you don't need calculus to show that there is a maximum in sin(t) + cos(t) - 1 at pi/4. A sketch or even intuition is enough, but formally: sin(t) + cos(t) = sqrt[(sin(t)+cos(t)^2)] = sqrt(sin^2(t) + cos^2(t) + 2sin(t)cos(t)) = sqrt(1 + 2sin(t)cos(t)) = sqrt(1 + sin(2t)), which has a maximum when 2t = pi/2. We need to be careful about using x = sqrt(x^2) if x might be negative, of course, but here we can see that it won't be in the domain 0
My guess for this maximum area problem geometrically is when the rhombus is a square or when one of the angles is a convenient 45°. Generally it will need an ingenious technique. That's how it usually works out in such problems.
This was my thought as well. I haven't actually worked out the angle though. Is our intuition correct?
So, it's easier without calculus actually. Having the eq. 2A = 2*sqrt{4-x²}-x², you take A as a constant, and you obtain a quadratic equation for x². For maximizing it you impose the discriminant is zero, so A= sqrt{2}-1. Then x² = 1- A, which is your result, and then replace its value in l, and terminado.
EXCELLENT journey!
Thx !
First the problem to solve these algebraically and/or geometrically is generally not easy!
Second, that is the power of the Calculus! So so so much easier with it!
Recall Kepler plodded for decades before he hit on his laws. If he had made any numerical or other mistakes he probably wouldn't have gotten his monumental results! Imagine if he had the Calculus a century or so later he would have avoided that drudery!
Once we know the value of x, the angle is easy to find by Law of cosines
Not even. You can just use 2arcsin on x/2 as seen in the triangle
@4:00 the square is also a rhombus ...
A square is a rhombus with equal diagonals...: As=x^2/2.
I took the challenge, solved with linear algebra. Haven't watched your approach yet. If I find that we solved it different ways, I will type up my solution since this was a really fun problem, at lease the way I took it.
From law od cosines is ((sqrt2)*l)^2=2-2cos alfa , max( sin alfa -l^2) for alfa=45 ,there is d/d alfa ( sin alfa -1+ cos alfa)=0, there is maximum
Without calculus, denoting theta as t here:
Following, from the one angle in the triangle being 135° and the opposite site being equal to 1, the area of a fourth of the shaded area is A = sin(t/2)sin(45°-t/2) / 2sin(135°) (Derived from sine rule A=ab x sin(135°) and the three angles being t/2, 135° and 45°-t/2) )
to maximize A we need to maximize the product sin(t/2) * sin(45°-t/2)
By symmetrie t/2 must be between 0 and 22,5° , but looking at slope of sin(x) clearly sin(t/2) is both smaller than sin(45°-t/2) and growing faster than sin(45°-t/2) is decreasing. Thus t/2=22,5° must hold to maximize the product.
(Is this without calculus, when I am argumenting with slopes? Maybe not, I am afraid, still found this a neat solution.)
Thank for Michael penn
You didn't have to find the square's side, since it's a rhombus itself and you know its diagonals. Nevertheless, awesome video.
Let’s call the points of the rhombus ABCD, and the other two points of the square EF (so now it’s BEDF)
let M be the center of AC, and theta be the angle MAB. a^2+(atan(theta))^2=1
(tan(theta)+1)a^2=1
a^2=1/(tan(theta)+1)
a^2*tan(theta)=tan(theta)/(tan(theta)+1)
a=1/sqrt(tan(theta)+1)
atan(theta)=tan(theta)/sqrt(tan(theta)+1)
Area of ABCD=2tan(theta)/(tan(theta)+1)
sqrt(2)l=2tan(theta)/sqrt(tan(theta)+1)
l=sqrt(2)tan(theta)/sqrt(tan(theta)+1)
l^2=2tan(theta)^2/(tan(theta)+1)
we want to maximize (2tan(theta)/tan(theta)+1)-2tan(theta)^2/(tan(theta)+1)
If tan(theta)+1>1, we can multiply by tan(theta)+1 and maximize 2tan(theta)-tan(theta)^2
We can let t be tan(theta)
We want to maximize 2t-t^2=2t(t-1)
This is a parabola with zeroes at 0 and 1 so its maximum is at t=1/2, which makes l be sqrt(2)(1/2)/sqrt(1/2+1)=1/2sqrt(3)
4:00 since all squares are rhombi you could have just used the rhombus area formula right??
The angle Theta is 45 degrees, right? (sin(Theta/2) = x/2, so Theta = 2 * arcsin(x/2))
Watching this video when I should be working, then he gives the best advice “I think this is a good place to stop”
Your videos are great
sin(pi/8)/sin(3pi/4)
I also have a challenge:
11:10 Represent √( 2 - √2 ) as (a - b*√2), where a>0, b>0.
Is it even possible?
If NO, explain, why.
If YES, calculate "a" and "b".
It's possible.
(a-bsqrt2)²=(a²+b²)-2absqrt2=2-sqrt2
a²+b²=2
ab=1/2
Therefore, a²+b²+2ab=(a+b)²=3=(a+b)²
Since a>0, b>0, a+b=sqrt3
If we see a,b as solutions of quadratic equation (x-a)(x-b)=0, we can say
2x²-2sqrt3×x+1=0
Then we can conclude x=(sqrt3-1)/2 or x=(sqrt3+1)/2
if a
@@김강민-j8x Thank yoy!
Your appoach looks still easier than mine.
But you made an important mistake:
The first equation is NOT "a²+b²=2"
but it should be "a²+2*b²=2" (as we square not "b" but "b*sqrt(2)").
I always did it like that:
1) a² + 2b² = 2
2) 2ab = 1
And I solved this system of equations
From the second equation, we have
2) b = 1 / (2a)
And we substitute this into the first equation:
1) a² + 2b² = 2
a² + 2 * [1 / (2a)]² = 2
a² + 2 * [1 / (4a²)] = 2 || * 4a²
4a^4 + 2 = 8a²
4a^4 - 8a² + 2 = 0 || :2
2a^4 - 4a² + 1 = 0
2*(a²)² - 4a² + 1 = 0
Substitute t=a² (t>=0):
2*(t)² - 4t + 1 = 0
And solving this quadratic equation for "t" we get:
Discriminant = DELTA = (-4)² - 4*2*1 = 16 - 8 = 8
sqrt(DELTA) = sqrt(8) = 2*sqrt(2)
t.1 = [4 - 2*sqrt(2)] / 4 = [2 - sqrt(2)] / 2
Since 2 > sqrt(2), there is [2 - sqrt(2)] > 0 and t.1 > 0, which is O.K.
or
t.2 = [4 + 2*sqrt(2)] / 4 = [2 + sqrt(2)] / 2
Since [2 + sqrt(2)] > 0, there is t.2 > 0, which is O.K.
So we get two cases:
case 1:
a² = t.1
a² = [2 - sqrt(2)] / 2
Since a>0:
a = sqrt{ [2 - sqrt(2)] / 2 }
Which is complicated... :(
Now I get another problem
HOW TO REPRESENT THE NUMBER "sqrt{ [2 - sqrt(2)] / 2 }" in the form of "c - d*sqrt(2)", where c,d>0?
And the second case:
case 2:
a² = t.2
a² = [2 + sqrt(2)] / 2
Since a>0:
a = sqrt{ [2 + sqrt(2)] / 2 }
Which is complicated... :(
Now I get another problem
HOW TO REPRESENT THE NUMBER "sqrt{ [2 + sqrt(2)] / 2 }" in the form of "e + f*sqrt(2)", where e,f>0?
intuition tells me instantly this area is maxed out when the rhombus is made of two equilateral triangles
ie: diagonal of the square is length 1 => L = sqrt(2)/2
now let's watch the video :D
Ya I started by noticing that this is like how the angle to a chord from the center is twice the angle from the opposite point. So I knew the acute angle of the rhombus was 45 degrees. Then it was simple.
7:25; when I had used the chain rule with that equation, it left me with -2x/2(4-x^2)^1/2, not -x^2/2(4-x^2)^1/2, so I believe that your answer is wrong.
I am studying geometrical drawing, this challenge is perfect for my study, thanks YT algorithm, and thanks Michael Penn for making me remember why I love calculus
My solution didn't use calculus and is *very* informal, so it's not a method I would recommend to others.
Consider the acute angle of the rhombus. If that angle is 0 then the area we're looking to maximize is 0, as that's the area of the whole rhombus. If that angle is 90, then that area is also 0, as now it's just overlapping the square. Obviously we want something in the middle, therefore the angle is 45 degrees.
From there, use the law of cosines to find the diagonal of the square: d^2=2-sqrt(2), which makes l=sqrt[2-sqrt(2)]/sqrt(2)
Forgot to simplify to sqrt(1-(1/sqrt(2)) whoops
Using θ, A = 4sin(θ/2)cos(θ/2) - 2sin(θ/2)²... crud, I gotta remember those double angle trig functions, because I'm pretty sure this reduces nicely for λ = θ/2.
Coudln't we just not split the problem diagonally (from down left to upper right) to simplify this problem to two triangles?
i thought you said that you were solving it without calculus
no, that is his challenge for _us_
maximum θ is arcsin(⎷(a-1/⎷2)), about 33°.
but the angle is 45 degrees... because it is 2 * arcsin( x / 2 ), or 2 * arcsin(sqrt(2 - sqrt(2)) / 2)
I loved this
And that's a good place... but he didn't stop because he used calculus.
"a rhombus is a parallellogram with all equal sides" is a partially redundant definition. "a rhombus is a quadrilateral with all equal sides" is better. You don't need parallellograms to define rhombi.
jokes on you I don't even know calculus
good place to stop
12:31
no homework for this video
@@aashsyed1277 6am EST I wake up early
it wasn't unlisted, it just showed up
@@aashsyed1277 yo calm dowm
@@aashsyed1277 stop screaming
@@aashsyed1277 OK!!!!!!!!!!!!!!!
@@aashsyed1277 all caps party
I literally don't understand how aash managed to comment so much on this video, chill dude
So "Good Place" at 12:30
I hope your builder didn’t hang your blackboard! Calculate the angle theta!
"And that's a good place t-" (* cuts off)
Well at 2.53 u said that diagonals of parallelograms intersect each other at right angles which is actually wrong since only the diagonals of square and rhombus intersect each other at right triangles but not that of others like rectangle and a simple parallelogram.
Solved it without calculus... but I get the wrong answer!
I tried it by using the Arithmetic Geometric Inequality Theorem. The idea is, the max area occurs when the arithmetic and geometric means are equal, right?
Let A = L and B = sqrt(2-L^2) - L
Then you're trying to get (A+B)/2 = sqrt(AB), where AB = the area of the shaded part. The answer works out nicely, is very easy to calculate... and is WRONG. Why?
EDIT: I think I understand my issue. Setting them equal will minimize the arithmetic mean (what I don't want), which is different from maximizing the geometric mean (what I do want). But I don't know how to fix it.
I just want to see some awesome infinite series and integrals here
I can’t even solve it with calculus
Theta=45
Hi I was wondering if you could show a solution to a problem of finding the radius of the small circle that is formed in the area between 3 larger but different touching circles please? Thus radius of small circle, r (the unknown) in terms of the large touching circles, R1, R2, & R3?
Sir, I feel there is fault during differentiation. While differentiating 1/(4-x^2)^1/2
When was that expression to be differentiated? What was the fault?
It’s correct
@@rbnn it's wrong :) if you know differentiation well, you will find the mistake at 7:15
@@suhail_69 1/2 comes from the x/2; 1/2 comes from the exponent; 2 comes from the x^2. It’s right.
@@rbnn you're right, sorry about that.
Use linear algebra
please make a video about how we dont know whether pi+e is rational or not, and how pathetic that is
x is no new variable. it's l times root 2.
I solved it with Trigonometry without Calculus. At the end it will need to find maximum value for sin (x) + cos (x) . It is my common knowledge the x will be 45 degree . but to prove , this is the way . 1/square root of 2 = 1/1.414 = sin 45 = cos 45 . Re-write equation by 1.414 * [cos 45 * sin x + sin 45 * cos x ] = sin (x) + cos (x) . then pull out this formula sin (x+y) = sin x cos y + cos x sin y . 1.414 * [cos 45 * sin x + sin 45 * cos x ] = 1.414 * sin(45 + x) . Maximum sin = 1 . then x =45 degree. Math is funny . I got theta first then work my way to get " L" . Micheal get " L" first without need to find theta.
Fact: he's using calculus but is challenging YOU to try it without calculus.
Angle Theta: 45.
The answer is approximately equal to 0.5412.
But derivatives are like the core of calculus?
Yes, but you can solve this problem without derivatives and therefore without calculus. It comes to a curve touching problem, which I showed a little way down the thread. It's a very insightful comment (a bit like Merry at the Moria-gate), as the "curve-touching" approach leads in to a whole wonderland of algebraic curve theory, which is pursued in the spirit of projective geometry, which eschews >>limits
Theta is pi/4?
yes
we can also use HERONS FORMULA to calculate the area of rhombus
I got tricked by the schedule, so people already posted the timestamp 🤷 No homework, that is.
What’s going on with the spam in the comments, though?
You are late this time 😂
@@_judge_me_not Yes, I know.
@@goodplacetostop2973 Well I deleted my comment
Coz it said that I can't find your's
I'm sorry I wrote the timestamp, people were asking
@@MathElite Don’t be sorry for that. It’s not a big deal
It would be much easier to solve with triangular function,.
hi, your calculation of As is a little 'complicated'...As is also un rombo => x*x/2 Great vidéo !!! Thanks !!!
It is x*y/2, not x*x/2, since it is possible that y =/= x.
@@megauser8512 Lol...in the case of a square (and A is a square)...x=y...
@@philippeillinger6287 Yeah, that makes sense.
shouldn't the existence of the figure be verified?
He sort of implicitly did that when he calculated the lower and upper bound for x.
Your problem statement doesn't say l has to be greater than 0. :)
You are making a big mistake of setting a square inside a rumbas. If you put the opposite vertices of the square on the opposite vertices of the rumbas, the other two vertices of the square are alternatively locating on the long diagonal of the rumbas. That is because the long diagonal of the rumbas is a perpendicular bisector of its smaller diagonal. Unless you are thinking of a Parallelogram and not a rumbas. So no matter how you draw your square, its other two vertices must fall on the long diagonal of the rumbas.
Dr. Zomorrodian, Mathematics professor
The title is "without Calculus" but you are using it!
The title, and so the challenge, was in fact for the viewer. But I guess it’s a miss, a lot of people have the same reaction as yours
That was easy
🥰🥰🥰...
You said solve "without calculus", then you used derivatives. Okay, I guess that's technically pre-Calc, but I assumed "without calc" meant without derivatives because who on Earth would have though to use integrals on this???
Without calculus is our HW. :)
He set the "w/o calculus" challenge for us. He was not bound by that constraint :)
And differentiation is definitively calculus, not pre-calc.
Great video, but he used calculus didn't he?
The ‘without calculus’ is a challenge for us viewers.
-> Challenge: Solve this without Calculus!
-> We solve a nice geometry problem using methods from first semester calculus.
That was a good place to stop this bs.
2=a jk