Question 577395
We know that


*[tex \LARGE A = 400 = 2\pi r^2 + 2\pi rh] (surface area of a cylinder). Also, the volume V is


*[tex \LARGE V = \pi r^2 h]


Several ways to do this. The easiest way is probably to use Lagrange multipliers, which states the maximum value of V (usually) occurs when


*[tex \LARGE \bigtriangledown V = \lambda \bigtriangledown A]


*[tex \LARGE \langle 2\pi rh, \pi r^2 \rangle = \lambda \langle 4\pi r + 2\pi h,  2\pi r \rangle]


Equating the vector components, we have

*[tex \LARGE 2\pi rh = \lambda (4\pi r + 2\pi h) \Rightarrow rh = \lambda (2r + h)] (1)


*[tex \LARGE \pi r^2 = \lambda 2\pi r ] (2)


(1) yields *[tex \LARGE \lambda = \frac{rh}{2r + h}] and (2) yields *[tex \LARGE \lambda = \frac{r}{2}]. Set these equal to each other:


*[tex \LARGE \frac{rh}{2r+h} = \frac{r}{2} \Rightarrow \frac{h}{2r+h} = \frac{1}{2} \Rightarrow 2h = 2r+h \Rightarrow h = 2r]


Thus, the maximum volume occurs when h = 2r, or when the height equals the diameter.


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Another solution that does not use multi-variable calculus. From the very first equation we have


*[tex \LARGE h = \frac{400 - 2\pi r^2}{2 \pi r} = \frac{200 - \pi r^2}{\pi r}]


Plug this into your expression for V, then find dV/dr.