Optimization problem using both cost and volume of a box.

Question

We are studying optimization, and she gave no notes on the topic. I  know now, from online videos, that I have to put variables in terms of each other to find what is asked, but this problem I am really puzzled on solving.

 

It states " a box manufacturer needs to build cylindrical containers with a volume of 6 ft^3. The material for the tops and the bottoms of the containers cost $2.00 ft^2, and the material for the side of the containers costs $1.50 ft^2. What is the cost of the least expensive container that can be built with the required volume."

 

So I approached it by figuring out what constitutes the cost.

The total cost = cost of Top/Bottom + cost sides.

I just have no idea how to fit the volume and cost together, or how to use some of this information.

Please help me, I am very confused.

Answer 1

Hi Olivia.

 

Cost optimization problems can be pretty tricky! What you'll notice is that the cost is a function of surface area. This particular cylinder has different costs for the sides and for the top, so you need to break the surface area of the cylinder into lateral surface area and surface area of the ends.

 

For a cylinder of radius r and height h, the lateral surface area is:

 

LSA = 2*pi*r*h [it's the perimeter times the height]

 

The top/ bottom surface area is:

 

SA = 2*pi*r²   [it's the area of two circles]

 

It says the cost of the side material is $1.50 per square foot, so the amount of money you'll spend on the sides is:

 

Cost/ lateral square foot * total lateral square feet

 

= $1.50*(2*pi*r*h) = 3*pi*r*h

 

The cost of the top and bottom will be

 

Cost/ end surface area * total surface area of the ends

 

= $2.00*(2pi*r²) = 4*pi*r²

 

So your total cost for 1 entire cylinder as a function of r and h is:

 

C(r,h) = 3*pi*r*h + 4*pi*r²

 

I don't like dealing in multiple variables though. It would be nice if everything were r. Fortunately, we have a volume equation to help us out with that. For a cylinder,

 

V = pi*r²*h

 

For THIS particular cylinder, V = 6. So:

 

6 = pi*r²*h

 

h = 6 / (pi*r²)

 

We can substitute this value into our cost function to make it entirely a function of r.

 

C(r) = 3*pi*r*[6 / pi*r²] + 4*pi*r²

 

C(r) = 18/r + 4*pi*r²

 

Cool. Now we have our cost function. We want to minimize this, so let's take its derivative and set it equal to zero.

 

C'(r) = -18/r² + 8*pi*r

 

0 = -18/r² + 8*pi*r

 

18/r² = 8*pi*r

 

r³ = 18/(8pi)

 

r³ = 9/(4pi)

 

r = (9/(4pi))^1/3

 

What an ugly value. Anyway, if this is your r_min then your Cost_min must be:

 

C = 18/(9/(4pi))^1/3 + 4*pi*((9/(4pi))^1/3)²

 

This result is roughly $30.18. That's how much the cheapest cylinder to make will cost to make. Usually the numbers are a lot prettier than this.

 

Hope this helps!