Let's just choose different values of price, which I'm writing as a little p,

and plug them into the function and see what the profit looks like.

And so in the table on this slide you can see I've

plugged in different values for price.

It's in the price column and I've used the equation,

the model to figure out what the profit is.

So if I charge $1.75 for

this product I actually don't make any profit off of it at all.

There's a negative profit, otherwise known as a loss, and of course, that makes

perfect sense, because 1.75 is less than the cost of production, which is $2.

If I were to price it $2, then I don't make any profit whatsoever because my

price is exactly equal to my cost.

So you get 0 for the second one and

then the subsequent numbers in there are just coming out of the profit equation.

Now if I look down through that table,

that optimization just corresponds to finding the biggest number in there.

And I've drawn a graph that shows you the profit as a function of price.

And you're already trying to figure out at which value of PDX axis

is the profit the highest.

Where's the top of that graph in other words?

So, this is a brute force approach because I haven't actually tried every value of p.

If I'm implementing this in a spreadsheet, spreadsheets have cells and

each cell you can only put in one number.

And so it's a discrete approach to solving this problem, and

it looks to me that the best value of price is somewhere sitting between 3 and

4, but I don't know exactly where between 3 and 4 it is right?

So, this gives me a sense of where the answer is.

And it might be fit for you, say it might be enough for you to say, I just want to

set the price between 3 and 4, but optimization does give us the potential to

be a bit more precise about it, so that's what I'm going to do now.

So, the calculus approached to these problems involves

the mathematical technique of differentiation and

what we need to be able to do is to find the derivative, which means the rate

of change of a function, with of the profit with respect to price,

and we need to see where that derivative equals to zero.

So optimization, the actual mathematics of optimization,

is not the goal of this course.

The goal of this course is to talk about modeling, and

this is one of the places that models are used.

And so I'm not actually going to do that, I'm going to present you with the results.

If you're interested in calculus one you can, and

its use in business you can certainly find other courses that will address that.

So I'm just going to skip to the answer here, it turns out that

by applying calculus to this problem you can obtain the optimal price.

And the optimal price which I'll write as Popt, opt for

optimal, is qual to c b / 1 + b,

where c is the production cost and b is the exponent in the power function.

So with this neat little mathematical model that we had for

quantity demanded as a function of price I'm able to leverage that equation,

leverage that model, and come up with an answer to the question.

What's the best price to set in order to maximize my profits?

Now, going back to this example, c was equal to 2,

that was the cost of production, and b was equal to negative 2.5.

If I plug in those numbers to the equation, you can convince yourself that

the optimal value for p, for the profit, is about equal to 3.33,

it's already three and a third is the best value for price.

So that is solution to the problem.

And by creating, or using a simple model for the quantity for

the demand I'm able to end up with a simple model,

you can even call it a rule of thumb if you want,

a simple formula for pricing.