Question 445838
We know that, from the constraint, h = 30 - r. Hence, the volume V is


*[tex \LARGE V = (\pi r^2)(30 - r) = -\pi r^3 + 30\pi r^2] If we suppose V is a function of r, we can take the derivative of V with respect to r:


*[tex \LARGE \frac{dV}{dr} = -3\pi r^2 + 60 \pi r]


The derivative is zero when r = 0 or r = 20. Clearly, r = 0 would imply V = 0. It can be checked that r = 20, h = 10 maximizes the volume, which is


*[tex \LARGE V = (20^2\pi)(10) = 4000\pi] cm^3.