Before I go on to explain optimization, I would like to make something very clear. Like I said by related rates, optimization and related rates are very different, and require different outlooks. In related rates, you are simply taking the derivative of a particular function that includes all the variables you are given. Here in optimization, your method will be taking the derivative of a function with only two variables. In related rates, all the variables change with respect to time; here, one will change with respect to the other.
The key to all optimization problems is to remember your ultimate goal what it is that you want to maximize or minimize. Everything must work toward that.
Well, how do you minimize or maximize something? Let us say you are trying to maximize the volume of a box. If I can give an equation that
represents the volume of the box, I can graph that equation. Recall from graphing that at the
maximums and minimums, the slope is 0. For
this reason, we take the first derivative of the equation, and set it to 0. This will give us every value at which the slope is 0, and it
will tell us where there are maximums and minimums to the equation. If that equation
represents the volume of the box, it will give us the
In these problems, before you take the derivative, you must have an equation with only two variables, such as x and y. This will require preliminary steps, as is detailed below. It is a crucial part of every optimization question.
Here is a question where you are asked to maximize the
You have 1350 cm2 of
material available. You would like to make a box, with a square base and closed top.
Find the height, width and length that would give the largest possible volume.
I will solve this problem soon. I have written it here merely to give you an idea of the the problems look like. First, I would like to write out the steps I will use to solve it.
This outline works for every optimization problem. It is possible to solve a problem without sticking precisely to the outline, but if you train with this outline, you will be able to use it unfailingly in problems that are otherwise impossible. Trust me on this; I've helped enough people to know.
Step 1 Find what you are trying to maximize or minimize. This will be stated explicitly.
Step 2 Write an equation for it. Use V = for volume and M = for material, A for area, etc. I must insist on using descriptive
variables, because in optimization, if you are sloppy, you lose track of what's going on.
Step 3 Try to get the equation into a two variable form, so you can take the derivative.
Step 3a To do step 3, you will often have to create a second equation from additional information given in the problem. This may require ingenuity, but it should become natural.
Step 3b Isolate one of the two variables in the
equation drawn in step 3a.
Step 3c In the original equation that you are trying to minimize or maximize, replace the variable you isolated in step 3b.
Step 4 Take the derivative of your two-variable equation.
Step 5 Set the derivative to 0, and solve for the value of the remaining variable.
Now back to our problem: "You have 1350 cm2 of material available. You would like to make a box, with a square base and closed top. Find the height, width and length that would give the largest possible volume."
Set up an equation for what we are trying to maximize the volume. Volume will equal height x width x length.
V = HWL
In this problem, the width is equal to the length, because I said it had a square base. So we can say:
V = HWL = LWL = L2H
Now we have only 3 variables, but we still cant take the derivative until we are down to two. At this point, when you are trying to merge them, you must scrutinize the problem for numbers that will help. The info has to be there. In this problem, I said that there is 1350 cm2 of material. Assuming that length and width are the same, because of the square base, the equation for the amount of material is:
(M =) 1350 = 4LH + 2L2 (These are the 4 side panels, and the top and bottom)
Now I am going to isolate one of the variables, and replace in the V equation the formula.
4LH = 1350 2L2 H = (1350 2L2)/4L
Now that I know what H is in terms of L, we can go back to the original equation and replace H.
V = L2H V = L2(1350 2L2)/4L = 1350L/4 L3/2
I now have only two variables left the V and L. I can now take the derivative. Notice that I am taking the derivative of the volume, which is the object I must maximize.
V = 1350L/4 L3/2 V = 1350/4 3L2/2
I will now set it to 0.
1350/4 3L2/2 = 0 3L2 = 675 L = 15
Now we are almost done. We know that at the maximum volume, the length is 15. Now we must plug that in to the earlier equations to find out what all the other dimensions are, including the volume.
W = L = 15
H = (1350 2L2)/4L = (1350 450)/60 = 15
Hmm, that makes a lot of sense, doesnt it? The best volume is gotten when the box is a perfect cube, with height, length, and width all equal! They are all 15. Now to find the volume:
V = 15 * 15 * 15 = 3375
Now you should have a rough idea of what is going on. I'll give another problem here, and I'm going to be a bit more brief, but you should be able to follow what I'm doing. As we solve it, refer back to the steps to see how precisely they are followed. Learn the method.
You want to make a can that holds 100 cubic inches of liquid. It is closed on the top and bottom. Using the most efficient dimensions, what is least amount of square inches of material you can use?
M = 2∏rh + 2∏r2 (the material for the side is the circumference times height, and the top and bottom are both standard areas)
That was the equation that I need to take the derivative of, because I am trying to minimize material. There are three variables, and I only want two. I am going to use another equation to solve that, using numbers from the problem.
V = 100 = ∏r2h h = 100/(∏r2) (the equation for volume was area times height)
M = 2∏rh + 2∏r2 = (2∏r)(100)/(∏r2) + 2∏r2 = 200/r + 2∏r2 = 200r-1 + 2∏r2
M' = 4∏r - 200/r2
I am setting it to 0. Setting this to zero and finding r is not easy, actually. To do this, you must combine the equation into a single fraction, and set the numerator to zero. This makes sense, because when the numerator is zero, the equation is zero.
= 4∏r - 200/r2 = 0 = ∏r - 50/r2 = (∏r3
- 50)/r2 ∏r3
- 50 = 0 r3 = 50/∏
And that's what the radius is. But the question asks that amount of material. So I'm going to have to figure out the height and work through the equation for material.
h = 100/(∏r2) = (100/∏)/(3√(50/∏)2 Now this looks real complicated, but there is a weird and interesting way to finish it. The first part, 100/∏ can be 2 * 50/∏, which is on the inside of the complicated part of the equation. you do the division, you actually are going to subtract exponents, 1 - 2/3, giving you and exponent of 1/3. So this comes out to:
2 * (50/∏)1/3 = 2 * r = 2r. (Remember that r = (50/∏)1/3)
Now we know the height is twice the radius. This is probably a general rule for maximizing cylinder volume, if I had my guess.
M = 2∏rh + 2∏r2 = M = 4∏r2 + 2∏r2 = 6∏r2 = 6∏[3√(50/∏)]2
That's the amount of material used! I was hoping for a simpler answer. Always remember when you are done with a problem to look back and wee what the question was. I might have forgotten altogether to go back and solve what the amount of material is, and you wouldn't have noticed.