# Thread: Finding the largest area??

1. ## Finding the largest area??

A rectangle is inscribed in the semicircle y = sqrt[4 - x^2]. Find its largest possible area.

I think I am suppose to take the derivative, but what do I do next??

2. Originally Posted by elpermic
A rectangle is inscribed in the semicircle y = sqrt[4 - x^2]. Find its largest possible area.

I think I am suppose to take the derivative, but what do I do next??
Center of the circle was (0,0)

We take a point (x,0) as one of the edge of rectangle

The edge just above (x,0) will have
y coordinate lying on the semicircle and given by

$\displaystyle \sqrt{4-x^2}$ , hence its coordinate is

$\displaystyle (x , \sqrt{4-x^2})$

--------------------------------
Make sure you get the above part before you continue.
---------------------------------

Area of this rectangle will be given by

$\displaystyle = 2x \times (\sqrt{4-x^2})$

...............{2 times of x because the length will lie on both side of the axis}

------------------------
if you don't get the above thing try drawing the diagram.
-------------------------------

This area needs to be max. so try finding the maximum
value of Area(x)

$\displaystyle A(x) = 2x\sqrt{4-x^2}$

for maxima

$\displaystyle \frac{d}{dx}A(x) = 0$

From above get the value of x for which A(x) is 0 . try this

value on A(x) to get max. area.

--------------------------
Be sincere enough to read below after you have solved.

Spoiler:

$\displaystyle \frac{d}{dx} 2x \sqrt{4-x^2} = 0$

$\displaystyle 2\sqrt{4-x^2}+\frac{-x^2}{\sqrt{4-x^2}} = 0$

$\displaystyle 2(4-x^2) =x^2$

$\displaystyle 8 = 3x^2$

$\displaystyle x=\pm \sqrt{\frac{8}{3}}$

Now find area(use +ve x).