An archive the questions from Mark's Spring 2019 Calculus I classes.

Optimization homework help

audrey

I’m having trouble with an optimization problem from MyOpenMath. My version of the problem looks like so:

A cylinder shaped can needs to be constructed to hold 500 cubic centimeters of soup. The material for the sides of the can costs 0.04 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.

I then need to find the radius, height, and minimum cost. Any suggestions?

mark

First, let’s find the objective function, i.e. C(r,h) which is the cost of the can as a function of the radius r and height h. The area of the side is 2\pi rh and, since it costs 0.04 cents per square centimeter, its cost is 0.04\times2\pi rh. Similarly, the areas of the both the top and the bottom are 0.06\times\pi r^2. Adding that all up, we get

C(r,h) = 2\times0.06\,\pi r^2 + 0.04\times2\pi rh.

To reduce the number of variables from two to one, we use the constraint, namely:

\pi r^2 h = 500.

Thus,

h=\frac{500}{\pi r^2}.

Plugging that into C(r,h) for h, we get

C(r) = 2\times0.06\,\pi r^2 + 0.04\times2\pi r \frac{500}{\pi r^2} = 0.12\pi r^2 + \frac{40}{r}.

Computing the derivative, we get

C'(r) = 0.24\pi r - \frac{40}{r^2}.

Setting that equal to zero and solving for r, we find that

r = \sqrt[3]{40/(0.24\pi)} \approx 3.75751.

Using that we can find both h and the minimal cost:

h \approx \frac{500}{\pi \, (3.75751^2)} = 11.2725.

C = 0.12\pi (3.75751^2) + 40/3.75751 = 15.968.