I have a question. Why can’t the area just be xy. Like would it still work. I tried it with radius 4 and I am getting 8 as my area. I don’t get why it has to be (2x) (2y) is there something else to do instead of splitting it into fourths
Hi! i have a question, why r^2 is a constant for this example? i saw time rates with the radius getting derived at, isn't it the same with optimization?
I also struggled with understanding why r is a constant. I think its helpful to consider that r is NOT a function of x and y. Because the rectangle is inscribed, by definition, we know r will not change even though there are infinite possibilities for x and y. Because of this, we treat it like a constant when deriving dA/dx
Why would you need to do all that? A square is a rectangle. You just have to draw a line (radius) from the center, let's call this A. Then, another line (radius) at 90° of the first line, Let's call this B. Now, you already have 2 sides of a triangle, just calculate the hypothenus using Pythagorean Theorem using the 2 sides above, sqrt(A^2+B^2). Because the 2 sides (A and B) are equal, same as the length of the radius (r), you can simplify this by doing sqrt(2r^2). Or you can alternatively, you can use the sine function, r/sin 45°, since the resulting triangle would be an isosceles triangle and you have a given 90° on 1 angle. The result would be 1 side of the rectangle. Since it is a square, then you just multiply it by itself. In simplest form, it would look like this, Area=sqrt(2r^2)^2 Alternatively, Area=(r/sin 45°)^2
your accent helps me pay attention i have no idea why
This video is going to save my final grade!! This solution makes sense now!!
Glad it helped. Good luck in your class.
Midterm exam tomorrow, I really appreciate this video
Good luck on your exam.
I have a homework question with the exact same same question lol, this was really helpful
calculus early transcendentals section 8th edition 4.7 question #25
@@gartyqam It's a very common Calculus book for college.
How did the r disappear at the end? Shouldn’t it be r-r/2^1/2?
it's like 1-1/2 = 1/2
I have a question. Why can’t the area just be xy. Like would it still work. I tried it with radius 4 and I am getting 8 as my area.
I don’t get why it has to be (2x) (2y) is there something else to do instead of splitting it into fourths
Because center of circle is at the origin.
@@crowsmathclass so is it necessary to split it up. My teacher taught us to have the area as xy
Yes. Because we center at origin.
@@crowsmathclass is there a situation where we can just use A= xy without 2x 2y
Like how would we know if it is centered and not centered
so are we solving for 2x at the end rather than singular x or y?? i was bit confused.
x
Can you please help me
So i was thinking what if the tye rectangle has a length of 11 cm and width of 6 cm.
How do we solve that ??
just input the numbers into x and y I think
How would you find the radius of the entire circle?
The radius is a variable. Hypothetically, you can plug any number to it.
Sir I am new to your channel and your explanation amazing it is clear as a crystal of water
Thank you for the nice comment.
Love your voice and this explanation is really easy to follow
big thumbs up
Thank you. Glad it helped.
Shouldn't the second term for the derivative of the area be -4x^2/sqrt(r^2-x^2) instead of -4x/sqrt(r^2-x^2)? -x*4x= -4x^2.
He did input it there unless I'm missing something? His answer is right.
Hi! i have a question, why r^2 is a constant for this example? i saw time rates with the radius getting derived at, isn't it the same with optimization?
We are looking for the dimensions of the rectangle. r is a constant in this problem. So we are looking for x and y.
I also struggled with understanding why r is a constant. I think its helpful to consider that r is NOT a function of x and y. Because the rectangle is inscribed, by definition, we know r will not change even though there are infinite possibilities for x and y. Because of this, we treat it like a constant when deriving dA/dx
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How can you say that area is maximum
As proof, use the 1st or 2nd derivative test. I think it will show you that that is indeed the maximum area
Very helpful, thank you!
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Thankyou sir!. This was pretty helpful.
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Why would you need to do all that?
A square is a rectangle.
You just have to draw a line (radius) from the center, let's call this A. Then, another line (radius) at 90° of the first line, Let's call this B. Now, you already have 2 sides of a triangle, just calculate the hypothenus using Pythagorean Theorem using the 2 sides above, sqrt(A^2+B^2). Because the 2 sides (A and B) are equal, same as the length of the radius (r), you can simplify this by doing sqrt(2r^2).
Or you can alternatively, you can use the sine function, r/sin 45°, since the resulting triangle would be an isosceles triangle and you have a given 90° on 1 angle.
The result would be 1 side of the rectangle. Since it is a square, then you just multiply it by itself.
In simplest form, it would look like this,
Area=sqrt(2r^2)^2
Alternatively,
Area=(r/sin 45°)^2
Super super helpful
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i love your accent
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Thank you, it really help me understand the problem.
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You should find second derivative and find Local maximum and minimum for area,which can give surety for maximum area
Thanks man. nice voice
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Thank you sir
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Thank u so much
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My god thank u so much
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Sir I am new to your channel and your explanation amazing it is clear as a crystal of water
Thank you sir
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