Calculus 1 question:

Find the area of the largest rectangle that can be inscribed in a circle of radius 1.

Question

Calculus 1 question: 

Find the area of the largest rectangle that can be inscribed in a circle of radius 1.

greatlife44 · Answer

$$x^{2} + y^{2} = 1$$

ace-n-it · Answer

k

greatlife44 · Answer

Here's the circle 

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greatlife44 · Answer

then I know that a rectangle we can express the area as 

l*w 

which in terms of the circle can be  xy = A

greatlife44 · Answer

@ganeshie8 @pooja195

greatlife44 · Answer

@mayankdevnani

greatlife44 · Answer

You see @surjihayer I also thought we could express the length like this: 

$$\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}} = \sqrt{4} = 2x $$

$$[\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}} = \sqrt{4} = 2y $$

A = 2x*2y

fortytherapper · Answer

I'm not completely sure, but I may have an idea

Since we are fitting a rectangle in a circle, then that if we draw a diagonal from the center to a corner, that would be the hypotenuse and that would equal the radius, right?

So if we created a triangle with that hypotenuse, maybe we can find the length some way? What would the length also be equal to?

fortytherapper · Answer

Actually, I should read replies before I actually start answering. I think you took the same path

greatlife44 · Answer

@fortytherapper @zelda  

here is what I was thinking

$$x^{2}+y^{2} = 1 $$

$$y^{2} = 1-x^{2}$$

$$y = \sqrt{1-x^{2}}$$


4xy = Area 


$$4x \sqrt{1-x^{2}} = Area $$

greatlife44 · Answer

then I think we need to get the derivative

fortytherapper · Answer

Would this derivative involve 2 product rules and a chain rule?

greatlife44 · Answer

Yeah we need to use a product and a chain rule... 

$$\frac{ dA }{ dx }(4x \sqrt{1-x^{2}}) $$

$$4x \sqrt{1-x^{2}}' + 4x' \sqrt{1-x^{2}} $$

greatlife44 · Answer

$$4x(0.5)*(1-x^{2})^{-1/2} * -x + 4 \sqrt{1-x^{2}}$$


Then we get this 

$$\frac{ -4x^{2} }{ \sqrt{1-x^{2}} } +4\sqrt{1-x^{2}}$$


$$-4x^{2}+4(1-x^{2}) =  -4x^{2}+4-4x^{2} = -8x^{2} = -4$$

$$\sqrt{ x^{2} }= \sqrt{\frac{ -4 }{ -8 }} = \frac{ 1 }{ \sqrt{2} }$$

then we would plug this back into our original equation 

$$4(\frac{ 1 }{ \sqrt{2} })*\sqrt{1-(\frac{ 1 }{ \sqrt{2} })^{2}} = 2 $$

greatlife44 · Answer

I asked my friend before, I didn't know how to set it up the right equation to find the area. but the rest isn't bad

fortytherapper · Answer

You're a genius, nice!