Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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Optimization Using the Second Derivative Test - Problem 2

Norm Prokup
Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.


The second derivative can be used to determine the minimum surface area of a cylinder with a given volume. Recall that the volume of a cylinder is V=πr2h, where r is the radius of the base, and h is the height of the cylinder. Also recall that the surface area of a cylinder is SA=2πrh + 2πr2(the circumference of the circular base multiplied by the height plus the area of the two circular bases). This is the equation we want to minimize.

Find the critical points of the surface area function by taking the first derivative. Then, by taking the second derivative, you can determine which of these critical points is a minimum (when the second derivative is positive, the point is a minimum).

Let's do another optimization problem. Manufacturers of Slurm want to make a cylindrical can that will hold 250 pi cm³ of a highly addictive beverage. Determine the dimensions that will minimize the amount of material needed for its construction.

So I've drawn a cylinder here. I need to introduce some variables. I'm going to call the height of the can 'h'. I'm going to call the radius of the can 'r'. So the volume of the cylinder, is the area of the base times the height. That's going to be Pi r², that's the area of the base, times the base is h.

So here is a problem where I have two variables, r and h. I have to try to get this in terms of one variable only. So I need to find a constraint, somewhere in the problem, that will allow me to define r in terms of h. It's this. The volume is 250 pi. This has to equal 250 pi. Then I realize, what am I actually minimizing here? The amount of material needed for its construction. That's area, and that's the surface area of the can.

So let me find an equation for surface area. The surface area is going to be the circumference times the height, so that's 2pi rh. That will give me the lateral area of the can. Then pi r² for the top, and pi r² for the bottom. That's plus 2 pi r². So that's the thing that I have to minimize.

Now this is a function of two variables. I need to replace h in this function. So let's take a look at our constraint. This is actually our constraint. I'm going to cancel the pi, and I have that h equals 250 over r². So let me substitute that into my area equation. So a equals 2pi r, and h is 250 over r² plus 2 pi r². Now r will cancel here, and we have 500 pi over r. Then we have plus 2pi r².

Let's think about what kind of domain we want for this. Now I have this volume constraint, 250pi, but I could really have any positive value for r even if it's very small. I could find an 'h' that's big enough to give me that volume that I need. So, as long as r is bigger than 0, I'll be okay. So that's going to b my domain r bigger than 0.

So I have my area function. Let's find its derivative. So let me copy up here first. 500pi over r plus 2pi r²; that's r for greater than 0. It's the derivative A'. This is really 500pi times r to the -1. So the derivative of that would be -500pi r to the -2 plus, and the derivative of this term is 4pi.

In order to find critical points, I'm going to need to get this into a single fraction. Right now, this represents a fraction that's being added to another term. They don't have the same denominator, let me show you what I mean. This is -500pi over r² plus 4pi r. So I need to get both of these terms with the denominator of r² in order to combine them.

Well, I could give this one a denominator of r² if I multiply the top by r². That will give me 4pi r³. Now I can combine them into -500pi plus 4pi r³ over r². Now I'm looking for critical points. I notice that this is undefined where r equals 0. It's not a critical point, because it's not in the domain.

So let me look at the numerator. When is this going to equal 0? A' equal 0 when 4pi r³ equals 500pi. So let me cancel the pi's. Divide both sides by 4. 500 over 4, what's that? Well, it's 250 over 2, which is 125. That's a perfect cube. So r equals 5, that's our critical point. So we have one critical point.

Now what I have to determine is that the second derivative is either always positive, or always negative. So I need to find the second derivative, A''. So looking at this, the derivative of this first term, the -2 comes out in front. I get +1000pi r to the -3. Then for this term, the derivative of 4pi r is just 4pi. Now all I need to know is, is this second derivative always positive or always negative? Well, my domain is r greater than 0. Now this is the same as 1000pi over r³ plus 4pi. Now this part's always positive. If r is always positive, this term will also be always positive. So A'' is always positive.

Now think about what a second derivative always positive means? It means the graph is always concave up. So our single critical point at r equals 5 is going to represent an absolute minimum. So A has an absolute minimum when r equals 5.

Now we were asked to find the dimensions, that will minimize the amount of material needed to make the can. R equals 5 is one of the dimensions. Now we need to find the height, h. Well, that comes from our constraint way back over here. This was our constraint. Remember we got it from setting the volume equal to 250pi. If I plug in my r equals 5 here, I get 25 in the denominator, 250 over 25 is 10. H is 10.

Now remember, the original 250pi that was in cm³, so all these lengths are in centimeters. So the final dimensions of the can are; a radius of 5cm, and a height of 10cm.

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