Since profit can't be negative, the production level that'll maximize profit is approximately equal to 220.
The cost function, C(x) is given by 12000 + 400x − 2.6x² + 0.004x³ while the demand function, P(x) is given by 1600 − 8x.
Next, we would differentiate the cost function, C(x) to derive the marginal cost:
C(x) = 12000 + 400x − 2.6x² + 0.004x³
C'(x) = 400 − 5.2x + 0.012x².
Also, revenue, R(x) = x × P(x)
Revenue, R(x) = x(1600 − 8x)
Revenue, R(x) = 1600x − 8x²
Next, we would differentiate the revenue function to derive the marginal revenue:
R'(x) = 1600 - 8x
At maximum profit, the marginal revenue is equal to the marginal cost:
1600 - 8x = 400 − 5.2x + 0.012x
1600 - 8x - 400 + 5.2x - 0.012x² = 0
1200 - 2.8x - 0.012x² = 0
0.012x² + 2.8x - 1200 = 0
Solving by using the quadratic equation, we have:
x = 220.40 or x = -453.73.
Since profit can't be negative, the production level that'll maximize profit is approximately equal to 220.
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