We have 100cm^2 of material to build a cylindrical can that will include the sides and bottom, but not the top. What are the dimensions of the can that will give the maximum enclosed volume?

Question

campbell_st · Answer

so 

$$100 = 2\pi rh + \pi r^2$$

then $$h = 50/(\pi r) - r/2$$.........................(1)

now volume of a cylinder is


$$V = \pi r^2h$$

so 

$$V = \pi r^2 (50/(\pi r) - r/2)$$

or $$V = 50r  - \pi r^3/2$$

find dV/dr

$$dV/dr = 50 - (3 \pi r^2)/2$$

let dV/dr = 0 and solve for r

$$r = \sqrt{100/3\pi}$$

Substitute into (1) above to find h

I think thats close to the solution

anonymous · Answer

I got it, thanks it actually solves quit nice in the end.