(Highschool) AP Calculus, Optimization

Question

A microwaveable cup-of-soup package needs to be constructed in the shape of a cylinder to hold 450 cubic centimeters of soup. The sides and bottom of the container will be made of styrofoam costing 0.04 cents per square centimeter. The top will be made of glued paper, costing 0.08 cents per square centimeter. Find the dimensions for the package that will minimize production cost.

Answer 1

Hi Phoebe,

Let's start by writing an equation that describes the total surface area of this cylindrical package. This is important because our cost, as you see in the word problem, is determined on a per-area basis for the different materials used. Also, this is to remind you of how the areas of the cylinder depend on its dimensions.

Let the cylinder have height h and radius R.

A_total = (area of top and bottom) + (area of sides)

A_total = 2(πR²) + h(2πR) = 2πR² + 2πRh

Now, we are also given a constriction on what the volume of this cylinder is going to be. Let's also write an equation for that because it will later be useful in writing a cost equation that is dependent only on one of the variables describing the dimensions of this can.

V = πR²h = 450

We can now start writing an equation for the total cost (C). This is the equation we will ultimately want to minimize for this problem. Notice how I am taking the different materials into account:

C = 0.04 (2πRh + πR²) + 0.08 (πR²)

Now it looks like we have an equation of cost that depends on 2 variables (R and h), which seems like bad news, but actually just as I said before, h and R are related by the volume constriction. Notice:

V = πR²h = 450

h = 450 / πR² (by rearrangement)

now, just plug this h into the above cost equation, and everything will be in terms of 1 variable, R. After you plug in and simplify (I skipped that here), the cost equation should look like this:

C = (36 / R) + 0.12 πR²

You're now ready to minimize this equation. Differentiate it with respect to R, then set that derivative equal to zero. When the derivative is equal to zero, the cost function will be at a minimum point (which is exactly what we are looking for)

dC/dR = 0.24 πR - (36 /R²) = 0

Solve this for R and you will get R = 3.6 cm (approximately). To find what h should be just plug this R value into the volume restriction equation:

V = πR²h = π(3.6)²h = 450

h = 11 cm (approximately)