dimensions homework

Question

A cylinder shaped can needs to be constructed to hold 450 cubic centimeters of soup. The material for the sides of the can costs 0.03 cents per square centimeter. The material for the top and bottom of the can need to be thicker, and costs 0.06 cents per square centimeter. Find the dimensions for the can that will minimize production cost.

Helpful information:

h : height of can, r : radius of can

Volume of a cylinder: `V = pi r^2 h`

Area of the sides: A = 2pirh

Area of the top/bottom: `A = pi r^2`

To minimize the cost of the can:

Radius of the can: __________________ 

Height of the can: __________________ 

Minimum cost: __________________  cents

Answer 1

Let's set up the two equations we can use to solve this problem.

450 = πr²h

C = 0.03A_s + 0.06A_tb

C = total cost

A_s = area of sides = 2πrh

A_tb = area of top and bottom = 2πr

Note: The reason the A_tb equation is multiplied by 2 is it is accounting for both the top and the bottom, which are the same.

We want to minimize cost, or "C" in our case. To find minimums of functions, we take the derivative and set it equal to zero. First, let's get our equation of "C" and have it be a function of one variable, let's say "r". We will use the volume equation to help us out with this.

C = 2πr(h + 1)

h = 450 / (πr²)

C = 2πr(450 / (πr²) + 1)

C = 900/r + 2πr

Now, let's take the derivative with respect to "r" and set it equal to zero.

dC/dr = -900/r² + 2π

0 = -900/r² + 2π

900/r² = 2π

450/π = r²

r = 11.97 cm

Plug this radius back into the original volume equation to solve for h.

450 = π(11.97)²h

h = 1 cm

Now, plug in r and h into the cost function to find the minimum cost.

C = 2π(11.97)(1 + 1)

C = $150.42