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.)

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 = ±

P"(x) = -6x

P() = -6 ()   < 0

by second derivative, 'P' max at    x = = 33.17 (rounds)

since x =   ,

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

Therefore, Maximum Profit

P() = 3300 - 50000 -

              = 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

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