The sum of the height and radius of a cylinder is 6cm. Find the dimensions of the cylineer that will have the largest volume. Find that largest volume (leave answers in terms of pi). The volume of a cylinder is given by V = (pi)(r^2)h.

Question

The sum of the height and radius of a cylinder is 6cm.  Find the dimensions of the cylineer that will have the largest volume. Find that largest volume (leave answers in terms of pi).  The volume of a cylinder is given by V = (pi)(r^2)h.

hartnn · Answer

h+r = 6
For largest volume,
V' =0
(equating derivative to 0 gives extrema(maxima or minima))
so, what is V' =.... ?

anonymous · Answer

V' = 2(pi)(r)(h)?

anonymous · Answer

@hartnn

hartnn · Answer

sorry, for late reply,
first use h =6-r and then you get V in terms of 'r' only,
only then differentiate.

stamp · Answer

$$V(r,h)=\pi r^2h$$$$r+h=6$$$$h=6-r$$$$V=\pi r^2(6-r)$$
$$V(r)=6\pi r^2-\pi r^3$$
$$V'(r)=12\pi r-3\pi r^2$$$$V'(r)=3\pi r(4-r)$$Solve V'(r) = 0
$$0=3\pi r(4-r)$$$$r=0, 4$$

We will exclude zero because that's a bogus dimension

So r = 4 means h = 2 (recall r + h = 6)

(r, h) of (4, 2) gives you your maximum value.

stamp · Answer

If we had supped r = 6 - h we would have$$V(h)=\pi (6-h)^2h$$$$V(h)=\pi (36-12h+h^2)h$$$$V(h)=36\pi h-12\pi h^2+\pi h^3$$$$V'(h)=36\pi -24\pi h+3\pi h^2$$$$V'(h)=3\pi (12-8h+h^2)$$$$V'(h)=3\pi(h-6)(h-2)$$Solve for V'(h) = 0

h = 6, 2

6 is a bogus dimension because that results in r = 0 so we go with h = 2, and that means

h + r = 6

r = 4

Again, (r, h) of (4, 2) gives us our max dimensions.

stamp · Answer

You see how solving for V(r) instead of V(h) is a cleaner approach