1. ## Maximum Area (optimization)

A rectangle is bounded by the x-axis and the semicircle
y = sqrt(25 - x^2)
What length and width should the rectangle have so that its area is a maximum?

I end up with A' = (-2x^2 + 25) / (25 - x^2)^1/2. Am I on the right track or did I make a terrible mistake?

2. Originally Posted by Archduke01
A rectangle is bounded by the x-axis and the semicircle
y = sqrt(25 - x^2)
What length and width should the rectangle have so that its area is a maximum?

I end up with A' = (-2x^2 + 25) / (25 - x^2)^1/2. Am I on the right track or did I make a terrible mistake?
You're on the right track...

However, I get

$\displaystyle A'(x)=\frac{2(25-x^2)}{\sqrt{25-x^2}}$

3. Originally Posted by VonNemo19
You're on the right track...
So I made A' equal to zero and... got stuck. I don't know how to proceed from there and isolate x^2. Any help would be appreciated.

4. Originally Posted by Archduke01
So I made A' equal to zero and... got stuck. I don't know how to proceed from there and isolate x^2. Any help would be appreciated.

$\displaystyle A'(x)=0\Rightarrow25-2x^2=0$

$\displaystyle x^2=\frac{25}{2}$

$\displaystyle \Rightarrow{x}=\frac{5\sqrt2}{2}$

This makes good sense because you would expect the greatest area to be given when the diagonal in the first quadrant makes an angle of $\displaystyle \theta=\pi/4$ with the x-axis. And the cosine of $\displaystyle \pi/4$ is...

$\displaystyle \sqrt{2}/2$.