I solved considering a quarter circle instead of a semicircle, and in that case you can easily see that the max rectangle (that is the half of the rectangle you are looking for) is when sinx*cosx is max. That means when the product between two numbers is max. And we know that the product between two numbers is max when they are equal. This means x =45 degree. CB=OB=4*sqrt2/2=2sqrt2 Then AB=2*2sqrt2=4sqrt2 Area=2sqrt2*4sqrt2=16
Thank you for your problem and solution. The area of the rectangle: It is less than 8pi. If M is the midpoint of AB , AM.AM = AB.AB /4 AM.AM = (a)(2r-a) ( products of intersecting chords) r = radius = 4 a is the distance of M below the semi-circumference a = 4- BC AM.AM = (4-BC)( 8 - (4 - BC)) A^2 = AB.BC. AB.BC where A is the area to be maximal. Substituting for AB= 2AM and AB.AB is 4AM.AM which is 4(4 - BC)(4+BC) A^2 = BC.BC( 4)(16-BC.BC) {When BC=4 AB would be 0 } Were BC = 3 A^2 would be 9.4.7 = 21 x 12 = 252 Were BC=2 A^2 would be 4.4.12 = 16 x 12 = 192 This square of the area will maximise when BC = Root (8) to be 256 AM.AM would be 16-8 and AB.AB would be 32. Validate by slight increases or decreases. Or prove by derived function equals zero at a stationary point. A^2 = 8 .4.8 Area of ABCD = 16 = (2 root (8) ) by (root (8)) ANSWER = 16
Let x be the length of the rectangle, then the width of the rectangle is equal to (√(64-x²))/2 and its area is equal to x*(√(64-x²))/2. This area is maximum if x²(64-x²) is maximum, and the function f(x)=-x⁴+64x² is maximum for x=4√2, and thus the maximum area of the rectangle is equal to 4√2*√(64-(4√2)²)/2=16.
El rectángulo más grande que puede inscribirse en un círculo de diámetro =8, es el cuadrado cuya diagonal es ese mismo diámetro ---> Máxima área ABCD =(8²/2)/2 =16 u². Gracias y saludos
👍 Let DC = x , BC = y then using Pythagoras theorem x^2/4+y^2 = 16 max value of (x^2/4)(y^2) = 8× 8 max value of xy/2 us 8 max value of xy is 16 hence max area = 16 square units
My hat goes off to you for such elegant solutions ! I particularly enjoyed the use of the AM-GM inequality : ) As for me, I used the derivative to determine the dimensions that result in the greatest area but I find the geometric route more enjoyable.
If the rectangle height is a and base b, then a² = 16-b²/4 A = ab A² = a²b²= (16-b²/4)b² = 16²-(b²/2 -16)² A² is maximum when b²/2 - 16 = 0 Hence Max A² =16² Max A = 16
rectangle height = r.sin θ rectangle widrh = 2.r.cos θ rectangle area = A(θ) = 2.r².sin θ. cos θ area will be maximum on θ in which dA(θ)/dθ = 0 dA(θ)/dθ = 2.r².(cos²θ - sin²θ) (cos²θ - sin²θ) have to be 0 therefore θ = 45° or π/4 back to rectangle area = A(θ) = 2.r².sin θ. cos θ rectangle area = A(θ) = 2.4².(½.2½).(½.2½) = 16 yay. it's kinda weird i missed this part of mathematics from my teenhood
I solved considering a quarter circle instead of a semicircle, and in that case you can easily see that the max rectangle (that is the half of the rectangle you are looking for) is when sinx*cosx is max. That means when the product between two numbers is max. And we know that the product between two numbers is max when they are equal. This means x =45 degree. CB=OB=4*sqrt2/2=2sqrt2
Then AB=2*2sqrt2=4sqrt2
Area=2sqrt2*4sqrt2=16
Thank you for your problem and solution.
The area of the rectangle:
It is less than 8pi.
If M is the midpoint of AB , AM.AM = AB.AB /4 AM.AM = (a)(2r-a) ( products of intersecting chords)
r = radius = 4 a is the distance of M below the semi-circumference a = 4- BC
AM.AM = (4-BC)( 8 - (4 - BC)) A^2 = AB.BC. AB.BC where A is the area to be maximal.
Substituting for AB= 2AM and AB.AB is 4AM.AM which is 4(4 - BC)(4+BC) A^2 = BC.BC( 4)(16-BC.BC)
{When BC=4 AB would be 0 } Were BC = 3 A^2 would be 9.4.7 = 21 x 12 = 252 Were BC=2 A^2 would be 4.4.12 = 16 x 12 = 192
This square of the area will maximise when BC = Root (8) to be 256
AM.AM would be 16-8 and AB.AB would be 32. Validate by slight increases or decreases. Or prove by derived function equals zero at a stationary point.
A^2 = 8 .4.8 Area of ABCD = 16 = (2 root (8) ) by (root (8))
ANSWER = 16
Let x be the length of the rectangle, then the width of the rectangle is equal to (√(64-x²))/2 and its area is equal to x*(√(64-x²))/2. This area is maximum if x²(64-x²) is maximum, and the function f(x)=-x⁴+64x² is maximum for x=4√2, and thus the maximum area of the rectangle is equal to 4√2*√(64-(4√2)²)/2=16.
El rectángulo más grande que puede inscribirse en un círculo de diámetro =8, es el cuadrado cuya diagonal es ese mismo diámetro ---> Máxima área ABCD =(8²/2)/2 =16 u².
Gracias y saludos
❤
👍
Let DC = x , BC = y
then using Pythagoras theorem
x^2/4+y^2 = 16
max value of (x^2/4)(y^2) = 8× 8
max value of xy/2 us 8
max value of xy is 16
hence max area = 16 square units
a*Sqrt(2)=8=> a=4*sqrt(2)=> Area of the rectangle = 0.5*(4*sqrt2)^2=16 sqr unit. ❤
My hat goes off to you for such elegant solutions !
I particularly enjoyed the use of the AM-GM inequality : )
As for me, I used the derivative to determine the dimensions that result in the greatest area but I find the geometric route more enjoyable.
If ∠BOC=θ, then BC=4sinθ, DC=2×4cosθ=8cosθ. ∴[ABCD] =BC×DC=32sinθcosθ=16sin2θ. Since 0
360°ABCD/180°=2ABCD (ABCD ➖ 2ABCD+2).
R = 4 cm
A = R² = 16 cm² ( Solved √ )
Maximum area is when ratio of rectangle sides is b=2h
I used calculus
OA = 4, radius
(4-x)^2+y^2=4^2 , from Pythagorean theorem
y = sqrt(8x - x^2)
A(x) = 2*(4-x)*sqrt(8x-x^2)
A'(x) = -2sqrt(8x-x^2)+2*(4-x)(8-2x)/(2sqrt(8x-x^2))
A'(x) = -2sqrt(8x-x^2)+(4-x)(8-2x)/sqrt(8x-x^2)
A'(x)=(2x^2-16x +2x^2-16x+32)/sqrt(8x-x^2)
A'(x) = 4*(x^2-8x+8)/sqrt(8x-x^2)
x^2-8x+8 = 0
(x-4-2sqrt(2))(x-4+2sqrt(2))
x = 4-2sqrt(2)
2*2sqrt(2)*sqrt(16-(4-x)^2)
2*2sqrt(2)*sqrt(8)=4sqrt(2)*2*sqrt(2)=16
x^2 + y^2 = 16, L= 2x . y = 2x (16-x^2)^1/2, diff. this equeton give x=V8, hence Lmax= 16
(a/2)^2 + b^2 = 4^2
(a/2 - b)^2 + ab = 16
ab = 16 - (a/2 - b)^2
Area = ab, maximised by a/2=b
Max area = 16
AD=y、OD=xとすると、
相加>=相乗より、xyが最大
になるのは、x=yの時であるから、x=y=2√2であり、その時の面積は、2√2×2√2×2=16
h=BC,A=2h√(16-h^2)...A'=0 per h=√8..Amax=2√8√8=16
長方形の縦をxとおくと、
横は2√(16-x^2)
ABCD=2x(16-x^2)^(1/2)=f(x)とおいて微分すると、
f'(x)=2(16-x^2)^(1/2)+2x(1/2)(16-x^2)^(-1/2)(-2x)=0となるxの値は、
2√(16-x^2)-2x^2/√(16-x^2)=0
16-x^2-x^2=0
8=x^2
x=2√2
ABCD=f(2√2)=2√2×4√2=16
If the rectangle height is a and base b, then a² = 16-b²/4
A = ab
A² = a²b²= (16-b²/4)b²
= 16²-(b²/2 -16)²
A² is maximum when b²/2 - 16 = 0
Hence Max A² =16²
Max A = 16
8×8÷4=16
16
rectangle height = r.sin θ
rectangle widrh = 2.r.cos θ
rectangle area = A(θ) = 2.r².sin θ. cos θ
area will be maximum on θ in which dA(θ)/dθ = 0
dA(θ)/dθ = 2.r².(cos²θ - sin²θ)
(cos²θ - sin²θ) have to be 0
therefore θ = 45° or π/4
back to
rectangle area = A(θ) = 2.r².sin θ. cos θ
rectangle area = A(θ) = 2.4².(½.2½).(½.2½)
= 16
yay.
it's kinda weird i missed this part of mathematics from my teenhood