Optimization of Revenue Calculus Question

Question

The weekly demand for the Pulsar 40-in. high-definition television is given by the demand equation

p = −0.06x + 569    (0 ≤ x ≤ 12,000)

where p denotes the wholesale unit price in dollars and x denotes the quantity demanded. The weekly total cost function associated with manufacturing these sets is given by

C(x) = 0.000003x³ − 0.03x² + 400x + 80,000

where C(x) denotes the total cost incurred in producing x sets. Find the level of production that will yield a maximum profit for the manufacturer. Hint: Use the quadratic formula. (Round your answer to the nearest whole number.)

Answer 1

Maximum profit means find the marginal profit, which is the first derivative of the profit function, P(x).

Let The Revenue = R(x), Cost =C(x) then the profit function, P(x) = R(x) - C(x)

But Revenue, R(x) = x*P(x) =x(-0.06x+569) = -0.06x² +569x

==> P(x) = R(x) - C(x) = -0.06x² +569x - (3x10^-6x³ - 0.03x² +400x+80,000)

= -3x10^-6x³ - 0.03x² + 169x - 80,000

Differentiating this function with respect to x, gives you the maximum profit function.

==>P^'(x) = dP(x)/dx = -9x10^-6x² - 0.06x +169

Now maximizing this functions means that, P^'(x) = 0

Using the quadratic formula, find x = (-b ± √(b² - 4ac))/2a = (0.06 ± √(9.684x10^-3))/(-1.8x10^-5)

==> x = -8,800.4 or 2,133.7. Pick the positive value of x = 2,134 as the level of production that will yield maximum profit.