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 = ± [tex]\sqrt{1100}[/tex]
P"(x) = -6x
P([tex]\sqrt{1100}[/tex]) = -6 ([tex]\sqrt{1100}[/tex]) < 0
by second derivative, 'P' max at x = [tex]\sqrt{1100}[/tex] = 33.17 (rounds)
since x = [tex]\sqrt{1100}[/tex] ,
recall that P(x) = 3300x - 50,000 - x³ from equation (A)
Therefore, Maximum Profit
P([tex]\sqrt{1100}[/tex]) = 3300[tex]\sqrt{1100}[/tex] - 50000 - [tex]\sqrt{1100} ^{3}[/tex]
= 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
