 Alright, let's take a look at another optimization problem. So again, generally speaking, the calculus is actually the easiest part of any optimization problem. The really difficult part is finding the objective function, which is something that is a topic from geometry, from algebra, and from precalculus. And again, the bad news is there is no formula that will tell you how to find the objective function. It cannot be done. And there's no good news. The only way to find the objective function is to read the problem carefully and apply all the tools of algebra and geometry that you know. So let's take an example here. We have a coffee shop that's going to sell coffee at $2 a cup, and it's going to sell 500 cups at that price, and we know something about how the sales will change if I should increase or decrease the price. And the question that the manager wants to know is should we raise or lower the price, because if we do that, maybe we'll increase revenue, maybe we won't, and let's take a look at it. And again, problems in the ideal world come with a question like find the maximum value of revenue, but in the real world we're not so accommodating, and so we have to infer that this actually is an optimization problem. And we can tell that because the question is asking us for a price to charge to maximize the revenue. So that says we're looking for the maximum value of some objective function, very specifically revenue, which is the product of the price per cup and the number of cups sold. So I can write the expression that way. And after some effort, I can figure out the number of cups sold could be modeled by 1500 minus 500x, so there's my revenue function. And I want to differentiate and find the critical point. So the derivative, it's a product, looks like that. And first off, the derivative is defined everywhere, so we don't get any places where the derivative is undefined. And the second place that I want to look at where is the derivative equal to zero. And so there's my derivative. I want it equal to zero, and I solve that equation x equals 1.5. And at this point, this is a graphing problem. I have the critical point x equals 1.5, and I want to find the sign of the first derivative, and that's going to be positive and negative after. And so my stick figure sketch of the graph looks something like that. And there's my revenue function, the maximum y value, the maximum value of the function occurs right here at x equals 1.5. And so that's where I want to set the price at. It gives me a maximum value. x equals 1.5 is an answer to the actual question. What price should we charge? But we might as well, while we're at it, include the actual revenue. If x equals 1.5, my revenue function tells me how to find what the revenue is. So I'll substitute that in and get a revenue of $1,025.