P(x) = -0.4x^2 + fx - m

Question

f represents desgin fee and m represents the montly office rent.

If $80 is charged for a design fee, and the monthly studio rent is 1,600; write an equation for the profit, P, in terms of x. Then how mch is the profit when 50 award designs are sold in a month? Finally, how many award designs must be sold in order to maximize the profit and show work algebraically and what is the maximun profit?

Answer 1

To get an equation for the profit, P(x), plug in values for f and m:

P(x) = -0.4x^2 + 80x - 1,600

The profit for 50 designs is:

P(50) = -0.4(50)^2 + 80(50) - 1,600 = $1,400

Now suppose we find where the profit equals zero:

-0.4x^2 + 80x - 1,600 = 0

This equation is in the form

Ax^2 + Bx + C = 0, which means we can use the quadratic formula:

x = [-B +/- (B^2 - 4AC)^(1/2)]/(2A)

x = [-80 +/- (80^2 - 4*-0.4*-1,600)^(1/2)]/(2*-0.4)

x = 22.54 or 177.5

The -0.4x^2 term in the original equation indicates that the graph of P(x) starts at zero, rises, levels out, then goes back to zero. We just found the two places where x equals zero so P(x) must be max halfway between them:

x = (22.54+177.5)/2 = 100 designs

Another way to find max profit is to find the vertex of the original equation.