Let x be the amount of money a company spends on advertising in millions and let P be the profit in millions of dollars, where P = 320 + 20x - 2x^2.

Question

a. What should they spend on advertising to maximize profit?

b. What is the maximum profit?

Answer 1

You could also take the 1st derivative of the function (slope) and set it to zero to find where x is a maximum.

-2x² + 20x +320

1st derivative is -4x + 20

-4x + 20 = 0

x = 5 ($5 million in advertising)

Then use the value x = 5 in the original function to find the profit at that level of advertising.

Answer 2

Tne maximum or minimum (as requires by the problem) for a quadratic is -b/a/2 where the function is given as ax² + bx+ c.

In this case, x= -20/(-2)/2 = 5. P(5) is the maximum value.

A. $5 million

B. 320+20*(5) -2*(5)²

= 320 + 100 -2*25 = 420 -50 = $370 million