Business Applications — Question 3

Question 3 - Solution

(a) Maximize Profit:

$$P'(x) = \frac{d}{dx}(-2x^2 + 40x - 120) = -4x + 40$$

Set $P'(x) = 0$: $$-4x + 40 = 0 \quad \Rightarrow \quad x = 10$$

Second derivative: $$P''(x) = -4 < 0$$ Hence, profit is maximized at $x = 10$.

(b) Maximum Profit:

$$P(10) = -2(10)^2 + 40(10) - 120 = -200 + 400 - 120 = 80$$

(c) Break-even Points (Profit = 0):

$$-2x^2 + 40x - 120 = 0$$

Divide through by 2: $$x^2 - 20x + 60 = 0$$

Solve using the quadratic formula: $$x = \frac{20 \pm \sqrt{400 - 240}}{2} = \frac{20 \pm \sqrt{160}}{2} = 10 \pm 2\sqrt{10}$$

$$\boxed{ \begin{aligned} &\text{Maximum profit occurs at } x = 10 \text{ units, } P_{\max} = 80 \text{ thousand dollars}, \\ &\text{Profit is zero at } x = 10 - 2\sqrt{10},\; 10 + 2\sqrt{10}. \end{aligned} }$$