Optimization Problems: Applications to Economics - Problem 1
Let’s solve an optimization problem that involves finding the maximum revenue. Find the number of x units that maximizes the revenue function r(x) equals 900x to the 2/3 minus 60x. What is the maximum revenue? So this problem actually asks for two things; the actual maximum revenue value and the number of units that maximizes revenue.
The first thing I like to do is find the feasible domain. What does that mean? The feasible domain is the domain for this function that makes sense in the context of this problem. This function may actually be defined for all real numbers. But even if it is, it might not make sense for all real numbers in the context of revenue.
Number of units produced for example, if you’re manufacturing some product, you’re not going to manufacture less than zero of them. So it makes sense for example to limit the domain to x greater than or equal to zero, for a start. But we want to know if there’s any maximum value that limits the domain here. Certainly x is going to be greater than or equal to zero to being with. We also want the revenue to be greater than or equal to zero. This formula won’t make sense if the revenue comes out negative.
Let’s see what values of x make it so that the revenue is actually greater than or equal to zero. Looking at the revenue function, I can factor out an x to the 2/3 and a 60. This 90 is 60 times 15, so if I pull out a 60, x to the 2/3, I’ll have 15 minus x to the 1/3. It’s really important to factor your function when you’re trying to figure out what this domain is.
Now here, looking at these two factors, this is the only one that’s capable of being negative. The reason I say that is x to the 2/3 is x to the 1/3 squared. Anything squared is going to be positive or zero. So this is the troublesome area. So in order to get an actual negative value, I’m going to need x to the 1/3 to be bigger than 15. So in order for this function to be defined, I don’t want to that to happen. So we need x to the 1/3 to be less than 15. That will happen if x is less than 15³. Let’s say less than or equal to. X is less than or equal to 15³, which is 3375.
This requirement will make it so that r(x) is positive or zero, not negative. This requirement is just so that that the problem makes sense. I’m not going to produce less than zero products. So my feasible domain is x is between zero and 3375. The advantage of having a closed interval domain, is that I can use a closed interval method for this problem. So that’s what I’m going to do.
I’ve rewritten the revenue function up here with its feasible domain. The next thing I have to do, no matter what optimization method I’m using, is find critical points. To find the critical points you need to take the derivative of the function. So r'(x). In the previous lesson we learn that this is the marginal revenue. So we’re finding where the marginal revenue equals zero or is undefined.
What’s the derivative of this? Well it’s 90 times the derivative so x to the 2/3. The 2/3 comes out in front. So 900 times 2/3 times x to the (2/3 minus 1) so -1/3. And then the derivative of the -60x is -60. Just looking at this you can see that this thing is going to be undefined when x is zero. The question is, are there any other places where we get a critical point? You just make that x equals zero is a c.p. It’s also an end point so my feasible domain so I’m going to check the value at zero anyway. I also want to find out where the derivative equals zero.
Let me set this equal to zero. 900 times 2/3 is 600. So this is 600x to the -1/3 minus 60. Add 60 to both sides, then I’m going to divide by 600. I get 1/10. X to the -1/3 is 1 over x to the 1/3. So I can take the reciprocal of both sides and I get 10 equals x to the positive 1/3. Cube both sides and I’ll get my value for x. X is 1000. This is my second critical point.
Now I’m going to need to use the closed interval method. The way the closed interval method works, is you make a table for your function r(x). You make a table of values. And you pick the points, the two endpoints of your feasible domain and any critical points on that interval, and you evaluate the function of each of those points. X, r(x), the two end points are zero, and 3375. And the critical point is 1000. So I need to evaluate the revenue function at each of these points.
Now if you look at the revenue function, when I plug in zero, I’m going to get zero. It makes sense. If I sell zero items; I’m going to get zero revenue. Now 3375, if I plug that in, it’s hard to see exactly what happens here. But if you recall this number came about because it made the cube root of x equal to 15. So this is actually a zero of this function that we already found.
You can check this on your calculator. If you plug in 3375, here you do get 0. So the remaining value to check is 1000. So x is 1000, you’re selling 1000 units. What happens then? Well 1000 to the 2/3 is 100. I’m going to get 900 times 100 minus 60 times 1000, 60,000. This is 90,000 minus 60,000, 30,000.
Now I was asked for two things, first the maximum revenue. This is it. Whenever you’re using the closed interval method, the biggest of these values is the maximum. And also when this occurs, it occurs when I sell 100 items. So the maximum revenue is 30,000 dollars and it occurs when x equals 1000. That means when I sell 1000 of my product.