Please take a look and see if I solved this math problem correctly. your feedback is very important.

5. You have been asked to design a 1-liter(1000cm^3) cylindrical can made with sheet aluminum.

what dimensions of the can(radius and height) will used the least total amount of aluminum?

hint 1. the aluminum must cover both ends of the can as well as the circular wall.
hint 2. the primary and secondary formulas are :

V=(pi)(r^2)(h)
S=2(pi)(r)(h)

......

im little confused because im not sure whats formula should i used for my primary equation and secondary.. when i get the first and second formulas set-up, i know how to substitute back to the primary equation, find the derivative, set the derivative equal to zero and find the critical numbers. can you show me the first two steps please.

The volume equation is useful because it allows you to solve for one of the variables in terms of the other.

You want V = 1000 so

(pi)(r^2)h = 1000 . You use this to get h in terms of r


h = 1000/((pi)r^2)

The material include you S equation, as well as two circles for end ends


Materials = 2(pi)(r)h + 2(pi)r^2

Here you use your (h = ...) equation to get rid of h

Materials = 2(pi)(r)(1000)/((pi)r^2) + 2(pi)r^2, which simplifies to

Materials = 2000/r + 2(pi)r^2

Since you want to minimize materials take the deriviation

Materials' = -2000/r^2 + 4(pi)r and set to 0


4(pi)r - 2000/r^2 = 0 Multiply both sides by r^2 to get

4(pi)r^3 - 2000 = 0

r^3 = 2000/(4(pi))

so r is the cube root of the right side

Then use your h = .... equation to get h

use the first formula,i.e. of volume so then u wud get an equation in two variables.