Barion Pixel Optimization problems in business | mathXplain

Contents of this Calculus 1 episode:

Economic problems, Revenue function, Function of product, Quantity function, Per-unit profit, Differentiation, Monotonicity, minimum, maximum.

Text of slideshow

Changes in the current price of a stock between 8:00am and 6:00pm are described by the following function, where at the xth hour of the day the stock price in thousand dollars is:

What was the opening and the closing price?
At which hour was the price the lowest and the highest?

At opening x=8, and at closing x=18, so we need to substitute these.

Now let’s see when the price was lowest and highest.

We need to differentiate for this:

Well, it appears that the maximum was at 8:00am, and the minimum was at noon.

The demand function of a product is:

where x denotes the unit price of the product. What unit price would ensure the highest income?

We have to differentiate again, but not this function, because this is the demand.

We need the revenues function.

Let’s see what the revenues is.

We have to multiply the quantity sold...

by the unit price.


And now we can differentiate.

The maximum revenue is at unit price 10.

Now, that we know all this, let’s try one last problem.

The per-unit profit of a product in dollars is:

where x is the weekly sales quantity, in 1000 units. What sales volume would result in optimal total weekly profit?

The per-unit profit means the amount of profit we make when we sell one product unit. Let’s say the per-unit profit is $5, and we sell 100 units. Then the total profit is $500.

The total profit is computed by multiplying the per-unit profit by the volume sold.


Let's see what the volume is.

Well, it is x, but given in 1000 units.

So, x=2 does not mean they sell 2 units, it means they sell 2000 units. If x=3, then they sell 3000.

So, the quantity sold is 1000x.


And now we can differentiate.

The maximum is at x=1, in other words, the weekly profit is maximized when 1000 units are sold.


We are left with x=0


Fractions always have to be changed to a common denominator before testing for signs!

It seems we are out of luck: there is no x-intercept.


We are not going to bother with this, it doesn’t worth it, let’s differentiate instead!

Due to x>3, we say good bye to this.


Never zero, it is negative for all x values

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