###### 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 1

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.

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Another common optimization problem is to determine the dimensions of a box so that the volume is maximized, given the surface area of some material. An open box can be formed by cutting out a square from each corner and folding the sides up--the goal of this problem is to find how large the squares should be in order to maximize the volume.

In this example, let the material be a square so that all sides are the same length--call it L. If the side of the square we want to cut out is x, the side length of the material after we cut out squares from each corner is L-2x. This will be the side length of the base of the box.

Remember that the volume of a box is the surface area base multiplied by the height. We know that the base surface area is A=(L-2x)2. The height of the box will be x, the side length of the square that are cut out. So, the volume of the box will be V=(L-2x)2x. You can find the maximum of the volume by first finding the critical points of this function by finding the first derivative, then evaluating the second derivative at the critical points. If the second derivative is negative, that critical point is a maximum.

Let's take a look at an example. We want to construct an open top box, with a square base, using only 75 square feet of material. Determine the maximum volume that the box can have.

So, here, I have a picture of the box. It's a square base box. I need to introduce some variables to start with. I'm going to call this dimension x. Since it's a square-based box, this dimension will also be x. I'll need a separate dimension, let's say y, for the height of the box.

The volume is going to be the area of the base, that's x², times the height, which is y. Now I have two variables here. I always want to try to express the quantity that I want to maximize or minimize, as a function of one variable. So I have to find some kind constraint or some other piece of information that will allow me to do that. It's this. It's the 75 square feet of material. So somehow I have to turn this into a constraint.

Well, 75 square feet of material means, that the total surface area of this box has to be 75 square feet. There are four vertical sides, and a bottom to the box. Each of the vertical sides has an area of x times y. So let me write that down. Area is 4 times xy. That's the area of the vertical sides. The bottom has an area of x². This all adds up to 75. So this is a relationship between x and y, that I can use to solve for y, and substitute into this equation and get a nice formula that's in terms of x only.

So 4xy equals 75 minus x². I want to solve this for y. So I'll divide this by 4x, and I'll get y equals 75 minus x² over 4x. Now substituting this back into my volume equation, I get x² times 75 minus x² over 4x. There's a little bit of cancellation that's going to happen. I lose an x here, and I lose an x here. This becomes x over 4 times 75. That's 75 over 4x. Then minus x² over 4 times x, -x³ over 4. That's my volume formula.

There is one other thing I have to think about, and that's domain. What domain does this formula make sense for? Let's take another look at the picture. Now first of all, it does make sense to restrict the domain so that x is at least positive, or at least no less than 0. So I would say x is greater than or equal to 0. There's also a maximum value for x. Remember there's only 75 square feet of material, so x², the base, has got to be less than or equal to 75. That means x has to be less that or equal to root 75.

Now I'm not going to worry about simplifying this. I just want to keep in mind that I have a close bounded interval domain. If I wanted to, I could use the closed interval method, because of this domain. I'd like to get some practice in with the second derivative test, and so let's do that method instead.

You'll find that this method is actually a very nice easy method, if the second derivative is easy to take. Now, I'm going to actually factor a 1/4 out of this. Actually I will factor out a 1/4x. Now I'll leave 75 minus x². We'll use this formula later when we're evaluating volumes. Let me differentiate it in this form. V' is going to be 75 over 4, and the derivative of this term is -3/4x².

Now we have to find critical points. So we find them by setting this equal to 0. That'll equal 0 when 75 over 4 equals 3/4x². Let me multiply both sides by 4/3. That will get rid of my 3/4. The 4's cancel, the 3's cancel leaving 25.

I have 25 equals x². So my critical points are +/-5. Well, -5 is not a critical point, because it's not in the domain. Where did I write that? Over here. I need to copy this up here. The domain is between 0, and root 75. So -5 is not in the domain, so I really only want +5.

Now I need to take the second derivative. The way the second derivative test works is, if the second derivative is always negative, then I'll have an absolute maximum at x equals 5. If it's always positive, I'll have an absolute minimum.

The second derivative is going to be, well the derivative of this is 0. The derivative of this is -3/2x. Now, keep in mind that my domain is between 0 and root 75. So this second derivative is always going to be negative. This value is going to be positive or 0 at the end point possible. That doesn't change the fact that this graph is always going to be concave down. The second derivative is negative. So at the one critical point, at x equals 5, we are going to have an absolute maximum.

So we have, by the second derivative test, an absolute max at x equals 5. That's not what we were asked. We were asked for the absolute maximum volume. So I need to plug that x equals 5 back into the volume formula. V(5) is 1/4 of 5. So 5/4 times 75 minus 5². 75 minus 25. That's 50. The 4's cancels with the 50 leaving a 2 here, and a 25 here. So that's 125 over 2, which is 62.5. The units of the volume since the length is in feet, area is in square feet, this will be cubic feet. 62.5 cubic feet is the maximum volume of the box.