 This last example of an optimization modeling problem is one that comes out of our textbook on page 285 it's number 43 a great business application of the calculus Here we have a chemical manufacturer selling sulfuric acid in bulk at a price of $100 per unit if the daily total production cost in dollars for X units is C for cost Equaling a hundred thousand plus fifty X plus point zero zero two five X squared and if the daily production capacity is at most 7,000 units How many units of sulfuric acid must be? manufactured and sold to Maximize profit. This is a great example of how you see calculus used in the business world here We're talking about maximizing profits of a company Anytime you want to get profit there is a set way to do that Profit is always going to be your total revenue. How much money total you take in? Minus the cost your production costs now that could include costs for production as we're told here It could include costs for maybe advertising Whatever all right here We just really have the production cost to deal with so we need to come up with an equation For the profit that we want to maximize as a revenue minus cost Revenue is how much total money you take in in this case We're told that the company is making a hundred dollars per unit So our revenue in this case is simply 100 times X a hundred dollars for each of the number of units they sell From that we need to subtract the cost equation they gave us now The nice thing about this is notice how we do have a primary equation that we want to maximize in terms of one variable So this one's pretty simple to proceed from here I've gone ahead and Re-written what we have for our profit equation I distributed the negatives how we were subtracting away the cost and I went ahead and simplified and Combined my similar terms so we have here a profit equation that we want to maximize of negative 100,000 plus 50 X minus point zero zero two five X squared We know that we have to take the derivative of that That simply is going to be 50 Minus point zero zero five Of course we have to do the point zero zero two five times two that's where the point zero zero five comes from Times X set that equal to zero again very easy to solve We end up with X is 10,000 and that's our critical number if there is going to be a maximum for this profit equation This is the value at which it's going to occur We've looked at examples in which we've used the first derivative test with a number line analysis we've seen a couple in which we've done a second derivative test and Substituting the critical numbers into the second derivative to determine if we have a relative maximum or a relative minimum This one I'm going to use the extreme value theorem. Remember that the daily production capacity is at most 7,000 units. So really we do have a domain here. We're assuming they're going to make more than zero Units right can't make a negative number of units so we really could think of the domain of this function as Being zero up to seven thousand because that's the most apparently that this Company can handle making in one day since we do have a closed interval for our domain We can very easily use the extreme value theorem to determine the location of our maximum We're going to set up a table of values We have our domain values in there zero and seven thousand and we also want to include our critical number of 10,000 that we had Remember that we substitute into our profit equation. Remember what that was after I simplified it You easily can do this on your table on your graphing calculator If you substitute the original profit equation under Y1 in your calculator, you can go into your table and We'll go ahead and complete our table of values. We have the endpoints of our domain Zero and seven thousand now think about what happens when they make no units I think it makes sense. They're not going to have any profits Because they're not making any money because they're not selling anything So I think the zero one makes sense. Obviously, that's not how they're going to maximize their profits by making nothing 7,000 units produces a profit of a hundred and twenty seven thousand five hundred dollars So that's possible. Let's check out what happens at ten thousand Which was our critical number when we set the derivative equal to zero that gives a hundred and fifty thousand Definitely looks better But think about the constraints of this problem The company's not capable of making ten thousand units So really they're stuck at seven thousand units until at least they're able to increase their productivity at their plant So in this case, this is a great example of what happens in real life They know that they could make more money if the if they could make and sell Ten thousand units, but they're only capable of doing seven thousand So that's really the best they're going to be able to do is really maximize and sell everything They are capable of making seven thousand units in doing so they have a profit then of a hundred and twenty seven thousand five hundred dollars