Let's just choose different values of price, which I'm writing as 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.

That'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 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 at $2, then I don't make any profit whatsoever because my

price is exactly equal to my cost.

So you get zero 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,

the 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 really trying to figure out at which value of p, the x-axis,

is the profit the highest?

Where is 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 in 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 three and four.

But I don't know exactly where between three and four it is.

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

And it might be fit for use.

It might be enough for you to say,

I just want to set the price between three and four.

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 approach 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, 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 the, I'm going to present you with the results.

If you're interested in calculus, well, and it's 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 P-opt,

opt for optimal, is equal to c times b over 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 2, that was the cost of production.

And b was equal to -2.5.

If I plug in those numbers to those equation,

you can convince yourself that the optimal value for

p, for the profit, is about equal to 3.33.

It's really 3 and a third, is the best value for price.

So that is the 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.