Application of derivatives to geometry

[Government$]

Homework Statement

Of all cylinders inscribed in sphere of radius R largest area of side(M) has cylinder which height is R\sqrt{2}. Prove.

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

 

[Dick]

Homework Statement

Of all cylinders inscribed in sphere of radius R largest area of side(M) has cylinder which height is R\sqrt{2}. Prove.

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

Your differentiation is wrong. You need to use the chain rule when you find the derivative of the second factor in your product rule.

 

[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

 

[Dick]

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

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

 

[Government$]

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