Business Applications — Question 8

Question 8

The weekly revenue (in thousands of dollars) for producing $x$ units is: $$R(x) = 120x - x^2$$ and the weekly cost is: $$C(x) = 40x + 200$$

• Determine the profit function $P(x)$.

• Find the number of units $x$ that maximizes profit.

• Compute the maximum profit.

Question 8 - Solution

(a) Profit Function: $$P(x) = R(x) - C(x) = (120x - x^2) - (40x + 200) = -x^2 + 80x - 200$$

(b) Maximize Profit: $$P'(x) = -2x + 80$$

Set $P'(x) = 0$: $$-2x + 80 = 0 \quad \Rightarrow \quad x = 40$$

Second derivative: $$P''(x) = -2 < 0$$ Hence, profit is maximized when $x = 40$.

(c) Maximum Profit: $$P(40) = -(40)^2 + 80(40) - 200 = -1600 + 3200 - 200 = 1400$$

$$\boxed{\text{Maximum profit occurs at } x = 40 \text{ units, } P_{\max} = 1400 \text{ thousand dollars}}$$