# Homework Help: Application of derivatives to geometry

1. Jan 13, 2013

### Government$1. The problem statement, all variables and given/known data Of all cylinders inscribed in sphere of radius $R$ largest area of side($M$) has cylinder which hight is $R\sqrt{2}$. Prove. 3. The attempt at a solution I understand how to prove this i only have problem with derivative: $M=2*r*Pi*H$ and $r=\frac{\sqrt{(2R)^2 - H^2}}{2}$ $M=Pi*H*\sqrt{(2R)^2 - H^2}$ $M^{'}=Pi*(\sqrt{(2R)^2 - H^2}) + \frac{Pi*H}{2*\sqrt{(2R)^2 - H^2}}$ Then $M^{'}=Pi*\frac{8R^2 -2H^2 + H}{2\sqrt{(2R)^2 - H^2}}$ so how can i from this get that $R*\sqrt{2} = H$ My textbook says that $M^{'}=Pi*\frac{4R^2 -2H^2}{\sqrt{(4R)^2 - H^2}}$ therefore $R*\sqrt{2} = H$ Last edited: Jan 13, 2013 2. Jan 13, 2013 ### Dick Your differentiation is wrong. You need to use the chain rule when you find the derivative of the second factor in your product rule. Last edited: Jan 13, 2013 3. Jan 13, 2013 ### Government$

So $M^{'}=Pi*(\sqrt{(2R)^2 - H^2}) + \frac{Pi*H}{2*\sqrt{(2R)^2 - H^2}}*(8R -2H)$

$M^{'}=Pi*(\sqrt{(2R)^2 - H^2}) + \frac{Pi*H*(4R - H)}{\sqrt{(2R)^2 - H^2}}$

$M^{'}=Pi*((\sqrt{(2R)^2 - H^2}) + \frac{H*(4R - H)}{\sqrt{(2R)^2 - H^2}})$

$M^{'}=Pi*((\sqrt{(2R)^2 - H^2}) + \frac{H*(4R - H)}{\sqrt{(2R)^2 - H^2}})$

$M^{'}=Pi*(\frac{(4R^2 + 4RH - 2H^2)}{\sqrt{(2R)^2 - H^2}})$

This time 4RH is making me a problem i feel like there is a better way to factor out this expression

4. Jan 13, 2013

### Dick

The derivative of (2R)^2-H^2 is just -2H. R is a constant!

5. Jan 13, 2013

### Government\$

I can't believe it. I spent almost 2 hours on this silly problem because of silly mistakes. Thank you.