Maximizing Profit

~A company is in business of manufacturing and selling luxury yachts. Since these are big ticket items, they estimate the maximum number they can manufacture per year is 100. Based on historical data points and trends, they have estimated that their profit in each year would be modeled by the following equation: P(x)= -2x³+210x²+7200x, where x represents the number of yachts sold. How many must be sold each year for maximum profit? (nearest integer)

~We can solve this problem two ways:
a) the derivative method
b) the calculator method

Let me show you both ways

a) The derivative method

We need to find values of x where a possible maximum occurs. These are know as critical values.

In this problem, P'(x) exists for all x since P(x) is a polynomial function. So, the only places to check would be where P'(x)=0 and the end point at x=100, the maximum number that will be manufactured.

So, P'(x)= -6x²+420x+7200. Setting P'(x)=0 gives the following

-6x²+420x+7200=0. Dividing both sides by -6 simplifies our equation and does not alter the solution, we get

x²-70x-1200=0. Checking the discriminant b²-4ac to see if it factors, we get

b²-4ac = √9700, not a perfect square. So, the quadratic formula is needed.

x=[-(-70)+√9700]/2 and x=[-(-70)-√9700]/2

We can reject the 2nd value since it is a negative.

We get, x=84.24 to two decimal places. Since the 2nd derivative at this value is negative, this gives a maximum point.

However, we need to check the value at x=100 and see if that gives a greater profit than x=84.24. This is where the calculator is very useful.

Code in the profit equation into the y= menu and go to the table of values (2nd graph).

Insert x=84.24 to get a profit value of 901170. Likewise, insert x=100 to get a profit value of 820000.

So, x=84.24 produces the greatest profit for x values between 0 and 100.

To check your answer, use

b)the calculator method

First, set up a friendly viewing window by looking at the table values for profit up to x=100.

All values are less than 1000000, so I let x_min=0, x_max=150, y_min=5000, y_max=1000000.

Any window size that gives you a decent view of the profit function in the given x range of our problem will work.

Now, press 2nd trace (calc menu), go to 4 (maximum), enter. You should see the graph and the maximum point. We need to confirm that the x value is x=84.24

You will be prompted for a left guess. I entered x=80, then a right guess. I entered x=100, then any guess between these. I entered x=90.

The actual value is given at the bottom of the window, x=84.24, the same as we got using the derivative method.

So, the company needs to sell 84 luxury yachts per year for maximum profit.