Probem38+Sabrina

A cone-shaped paper drinking cup is to be made to hold 27 cm^3 of water. Find the height and raduis of the cup that will use the smallest amount of paper.

Just a little clarification here...what are you trying to do, minimize or maximize, and what is it that you are trying to do this to? What is your objective function?

V = (1/3)(pi)(r^2)h 27 = (1/3)(pi)(r^2)h 81/[(pi)(r^2)] = h

A = (pi)r * sqrt[(r^2) + (h^2)] A = (pi)r * sqrt[(r^2) + (81/[(pi)(r^2)])^2] A = (pi)^2 * r^2 * [r^2 + (6561/[pi^2 * r^4])] Did you square both sides here (which would be fine, as long as you remember to square the A also...and then realize that you will be doing implicit differentiation (which is not a big deal)) A = [(pi)^2 * r^4] + 6561 Fix right here -- this is not correct! I am assuming that you are distributing π r² over [r²+ (81/ π r²)²]

take the derivative: which will of course be different if you fix the problem above. A'=4(pi)^2 * r^3

WHAT IS THE RADIUS?

Kevin Wu: Let's find the interval first: The cone can be extremely long and have an extremely small radius, say 0. The cone can also be extremely short and have an extremely large radius: since volume is 27cm^3, the largest radius can be no more than A=pi r^2=27 r=(27/pi)^.5 Set derivative equal to zero to find critical points/possible answers: 0=4(pi)^2 * r^3 r=0 (can't be an answer, so the answer must be one of the end points)

Look at the restrictions we are given. Our restriction involves volume, in which r and h are related inversely (as r gets smaller, h must increase in order to preserve the volume of 27). So I agree that r can be very small, even almost 0, but that results in h being astronomically big -- which is still a possibility! What happens to r when h gets small? It might be better to examine r in this context to find its greatest value.

hmm... I am not sure if my reasoning is correct, but I think the answer might be r<(27/pi)^.5. (The answer has to be that endpoint, but not necessarily that value)

I really do not know how to solve this problem.

We are trying to minimize surface area of the cone SA= pi*r*sqrt(r^2 + h^2) A= 1/3 pi*r^2*h

27 = 1/3 pi*r^2*h 81/(pi*r^2) = h

substitute into SA: SA = pi*r*sqrt( r^2 + ( 81/(pi*r^2))^2 )

take the derivative: SA' = pi * sqrt( r^2 + (  81 / (pi*r^2) ) ^2 ) + [pi * 2r^2 * (-26244 / pi*r^-5)] / 2* sqrt( r^2 + ( 81 / (pi*r^2)) ^2 )

...this can't be right...

Kevin: Alright, I tried to redo this problem, and this is what I have: