A microwaveable cup-of-soup package needs to be constructed in the shape of cylinder to hold 500 cubic centimeters of soup. The s...

Question

A microwaveable cup-of-soup package needs to be constructed in the shape of cylinder to hold 500 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 0.05 cents per square centimeter. Find the dimensions for the package that will minimize production cost.

Helpful information:

h = height of the cylinder, r = radius of the cylinder

Volume of the cylinder: V = pi * r² * h

Area of the sides: A = 2 * r * h * pi

Area of the top/bottom: A = pi * r²

To minimize the production cost:

Radius = ______ cm

Height = ______ cm

Minimum cost = _____ cents

please explain, I'm very lost on what I should be doing for the other 9 very similar homework assignments as well and would like to understand if I can.

Answer 1

So first thing with these types of optimization problems, you want to create an equation that will give you the Cost, as that is the final objective of these problems.

Step 1: we create a cost function C(x) that will add up all the pieces of the soup package and how much it costs to make each piece. So if the soup is a cylinder, we have the bottom and the side of the cup made of styrofoam, and the top circle made of paper. The cost of each material will usually be given to you in the problem.

C = 0.04*bottom + 0.04* side + 0.05*top

Step 2 : We need to actually find the equations that describe area of the bottom, top, and side. Thankfully those are also given to you. The side is just the formula for area of a rectangle, and the top and bottoms are areas of circles. So replace those formulas into C.

C = 0.04 (π r² ) + 0.04 (2π rh ) + 0.05 (π r²)

Distribute and combine any like terms that you can.

C = 0.08π rh+ 0.09π r²

We now have our cost function! Because this is calculus, we can infer that we will have to be taking derivatives to find a minimum cost. But how are we supposed to do anything with this if it has two different variables, r and h?

Step 3: We need to find some equation that has both r and h in it and try to replace it into the cost function. You were actually given that equation: its the V= π r² * h volume of a cylinder, and we are told we must have a cup with 500cm³ volume, or

500= π r² * h

We can use this constraint to make the Cost function a function of only one variable. Write this equation in terms of h, meaning the h should be by itself. Divide π r² from both sides to get

h = 500/ π r²

Step 4: Now we can replace the h variable in the cost function with 500/ π r²

C = 0.08π r(500/π r²) + 0.09π r²

See how there are only r s now? Now we can simplify this and do some calculus!

C = 40/r + 0.09π r²

Step 5: Find the minimum by taking the derivative of this function and find the critical values. This is where you set the derivative equal to zero.

C' = 0.18π r -40/r²

0.18π r -40/r² = 0

So, doing some algebra, you should see that we have two possible values of r, r = 0 and r = (40/0.18π)^1/3

Because we have no interest in making a cup with 0 radius, thus 0 area, we look only at r = (40/0.18π)^1/3 to be our radius that will give us the biggest cup for our cost.

To find the corresponding minimum height, just substitute r into that volume equation and solve for h.

h = 500/ π ((40/0.18π)^1/3)²

h = 9.3 approx

So a cup with a height of 9.3 and a radius of (40/0.18π)^1/3 gives you the most area for the least amount of material. Follow these steps as guidelines for some of your other problems!

1. Create a cost function using the different materials and pieces needed. Draw a picture and label it!
2. Insert any given geometric formulas into the cost function (this function will now likely have two variables, in this case it was r and h)
3. Use a formula with both variables in it and substitute it in (for us it was the Volume formula = 500, we sometimes call this the constraint function, because we are constrained by the problem telling us the cup must have a volume of 500). You should have a cost function with one variable when done.
4. Take the derivative of this cost function, set equal to zero to find the critical points.
5. Substitute your critical point back into the constraint function to find the other variable.

Answer 2

desmos.com/calculator/dmyuvd5epy

I failed to mention in the video that the 1st or 2nd derivative test can be applied to show that the determined value of r does indeed generate a minimum cost (as opposed to a max).