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 = ± 
P"(x) = -6x
P( ) = -6 (
) = -6 ( )   < 0
)   < 0
by second derivative, 'P' max at    x =  = 33.17 (rounds)
 = 33.17 (rounds)
since x =   ,
 ,
recall that P(x) = 3300x - 50,000 - x³ from equation (A)
Therefore, Maximum Profit 
P( ) = 3300
) = 3300 - 50000 -
 - 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