Business Applications — Question 1

Question 1 - Solution

(a) Profit is revenue minus cost: $$P(x) = R(x) - C(x) = (50x - 0.2x^2) - (10x + 100) = 40x - 0.2x^2 - 100$$

(b) Maximize $P(x) = -0.2x^2 + 40x - 100$.

This is a downward-opening parabola. The maximum occurs at: $$x = \frac{-b}{2a} = \frac{-40}{2(-0.2)} = \frac{-40}{-0.4} = \boxed{100}$$

(c) Maximum profit: $$P(100) = 40(100) - 0.2(100)^2 - 100 = 4000 - 2000 - 100 = \boxed{1900}$$

Answer: $$\boxed{\text{Max profit occurs at } x=100 \text{ items and the maximum profit is } \$1900.}$$