Advertising expenditures in thousands of dollars x have been

Advertising expenditures in thousands of dollars, x, have been found to relate to profit, P, approximately as

P = x3 - 100x 2 + 3125x .

What advertising expenditure would produce the maximum profit, and what profit is expected at this expenditure?

P(x) = x^3 - 100x^2 + 3125x

P\'(x) = 3x^2 - 200x + 3125

Solving the quadratic equation

3x^2 - 200x + 3125 = 0

x = 41.6666, x=25

P\'\'(x) = 6x - 200

P\'\'(25) = 6(25) - 200 = -50

Hence the function is having at maximum at x=25

P(25) = (25)^3 - 100(25)^2 + 3125(25)

=> 31250 thousand