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

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

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

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Make sure you get the above part before you continue.
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Area of this rectangle will be given by

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

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

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if you don't get the above thing try drawing the diagram.
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This area needs to be max. so try finding the maximum
value of Area(x)

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

for maxima

$
\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.

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Be sincere enough to read below after you have solved.

Spoiler:

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

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

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

$
8 = 3x^2$

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

Now find area(use +ve x).