Question

A manufacturer estimates that its product can be produced at a total cost of C(x) = 50,000 + 100x + x3 dollars. If the manufacturer's total revenue from the sale of x units is R(x) = 3400x dollars, determine the level of production x that will maximize the profit. (Round your answer to the nearest whole number.)

Answer 1

Answer:

The level of production x that will maximize the profit is: 22,966

Explanation:

C(x) = 50,000 + 100x + x³

R(x) = 3400x

P(x) = R(x) - C(x)

      = 3400x - [50,000 + 100x + x³]

      = 3400x - 50,000 - 100x - x³

      = 3300x - 50,000 - x³   .................... (A)

P'(x) = 3300(1) - 0 - 3x²

       = 3300 - 3x²

At a critical point, P'(x) = 0

∴   0 = 3300 - 3x²

  3x² = 3300

    x² = 1100

     x = ± $\sqrt{1100}$

P"(x) = -6x

P($\sqrt{1100}$) = -6 ($\sqrt{1100}$)   < 0

by second derivative, 'P' max at    x = $\sqrt{1100}$ = 33.17 (rounds)

since x =  $\sqrt{1100}$ ,

recall that P(x) = 3300x - 50,000 - x³ from equation (A)

Therefore, Maximum Profit

P($\sqrt{1100}$) = 3300$\sqrt{1100}$ - 50000 - $\sqrt{1100} ^{3}$

              = 3300(33.17) - 50,000 - 33.17³

              = 109461 -50,000 - 36495.26

              = 22,965.74

Maximum profit is 22,966 to the nearest whole number