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Optimization Problems: Applications to Economics - Problem 2

Teacher/Instructor 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.

To optimize profit, first recall that profit is revenue minus cost, or pi(x)=r(x)-c(x). Also, recall that the domain of this function will be where x is greater than or equal to zero (because you cannot look at the profit from a negative number of items). Then, take the derivative of this function (recall the sum rule of derivatives, so pi'(x)=r'(x)-c'(x)). Then, by finding the critical points and the first or second derivative test, you can determine which critical point is a maximum.

Let’s take a look at another optimization problem, this one involving optimizing profit. Find the price p per unit that maximizes the profit function, I’ll call it pi(x), if cost c(x) is 12000x plus 26000. And the demand is described by the equation p, which is the price, equals 30000 -2x². So there’s a lot going on in this problem.

First of all the function to be maximized is not present here, I’m going to have to conjure it up. I need to know that profit is revenue minus cost. That’s an important fact from Economics. And I also need to figure out what exactly the revenue because it’s not given to me directly. So let me do that first.

Revenue, is often written R(x) and it’s price times the number sold, the price per item times the number of that item sold, it makes sense. And then this price I’m given is 30,000 minus 2x². When I multiply this through I get 30,000x minus 2x³. This is the revenue.

The profit is going to be revenue minus cost. So profit, which I’m calling Pi(x), and I’m dong this by the way and not calling it p(x) because I already have a p for price. So pi(x); revenue minus cost. Basically money in minus money out. Revenue is 30,000x minus 2x³. And cost is 12,000x plus 26,000. I'm going to put parenthesis around this (12,000x+26,000).

Let me simplify this. I have a -2x³; I’ll pull that out in front. A 30,000 x minus a 12,000x, that’s plus 18000x, and then I have a minus 26,000. So this is my profit function. The only thing that remains to be done is to find the feasible domain. I’m going to say let’s make our feasible domain just x greater than or equal to zero. First of all, it doesn’t make sense for x to be negative. X is the number of items manufactured and sold, so it doesn’t make sense for it to be negative. So this is very reasonable.

There might be some further restrictions on x but I’m not going to find them. Not finding them means, I can’t use the closed interval method, I don’t have the domain which is a closed bounded interval. However, I still have two other methods that I can use; the first derivative test or the second derivative tests. I’ll use one of those two. So feasible domain, and this is my profit function. Let me copy this up on the right side of the board. Pi(x) equals -2x³ plus 18,000x minus 26,000 and x is greater than or equal to zero.

The next thing I have to do is find critical points. For that I need the derivative. Pi'(x). The derivative is going to be -6x² plus 18,000. The derivative of this part is zero.

Now critical points are where the derivative is either undefined or equal to zero. Now this is never undefined, but it can equal zero. So I’m going to solve this equation. I get, I pull this 6x² over to the right side, I get 18000 equals 6x², and then divide both sides by 6. I get 3000. 3000 equals x².

That means x is either plus or minus root 3000. One of these answers doesn’t make sense, not at least with our feasible domain. Our feasible domain is x is greater than or equal to 0. The negative answer is not going to be a critical point, because it’s not actually in the domain. Critical point is x equals the square root of 3000.

First derivative test. How does that work? We’re going to make a number line. And we’re going to check to see where Pi', the derivative of this function, is positive, negative or zero. So let me write Pi'(x). We’re only concerned with x greater than or equal to zero, and we shall also mark the root 3000, that one critical point. I know that Pi' is zero at root 3000, that was what I just found out. I just need to determine whether it's positive or negative here and positive or negative here. So I can check points.

In fact I can check the easiest point, zero. Pi' is going to be 18,000 at zero which is positive. So it’s positive all the way up to root 3000. What happens afterwards? Well let's try a bigger number than root 3000. Let’s say like the square root of 10,000 which is 100. That’s bigger. So I’m going to use 100. If I plug 100 into this function, 100² is 10,000 and -6 times 10,000 is -60,000. -60,000 plus 18,000, doesn’t really matter it’s clear that it’s negative. -60 plus 18, definitely negative. So always negative over here.

What I have here is the profit function pi is always increasing up to this point, and always decreasing afterwards. That means I have a maximum right here when x equals root 3000. Max profit occurs when x equals root 3000, by the first derivative test for absolute max and min. Remember for that test you have to check that the first derivative is positive over the entire domain from zero all the way up to root 3000, and then negative from root 3000 afterwards. Our check guarantees that.

The other thing I need to do is just make sure did I answer the right question that they ask in the problem? I found out what x value makes the maximum profit but I don’t think that’s what was asked. Let me check back to the problem here. It says find the price p per unit that maximizes profit. The only relationship I have here between p and x, which is what I found, is this one. So I’m going to use that formula to convert my x value into a p value. Let’s do that right now.

Price p that maximizes profit will be 30,000 minus 2x², 2 times root 3000². Root 3000² is 3000. Twice that is 6000. So 30,000 minus 6000 is 24,000. So whatever item we’re selling, maybe it’s cars, 24000 dollars is the price that will maximize profit.

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