The cost function for a certain company is C = 40x + 800 and the revenue is given by R = 100x − 0.5x2. (THIS IS TO THE SECOND POWER) Recall that profit is revenue minus cost. Set up a quadratic equation and find two values of x (production level) that will create a profit of $800.

  x = {40, 80}

Step-by-step explanation:

The profit equation is ...

  P = R - C = 100x -0.5x^2 -(40x +800)

  P = -0.5x^2 +60x -800

We want to find values of x such that P = 800

  800 = -0.5x^2 +60x -800

  1600 = -0.5x^2 +60x . . . . add 800

  -3200 = x^2 -120x . . . . . . . multiply by -2

  400 = x^2 -120x +3600 . . . . add 3600 to complete the square

  400 = (x -60)^2 . . . . . . write as a square

  ±20 = x -60 . . . . . . . . . take the square root

  x = 60 ± 20 = {40, 80}

The values of production level (x) that will create profit of $800 are 40 and 80 units.

29.5977692753

step-by-step explanation:

this linear equation would have just one solution.

step-by-step explanation:

the reason why this would have one solution is that it is an expression with just one variable. since that variable is not raised to a power, it will always have just one solution.