1. **State the problem:** We want to find the order size $x$ that minimizes the cost per unit, given the total cost function $$C(x) = 5x^2 + 320.$$ 2. **Define cost per unit:** The cost per unit, $P(x)$, is the total cost divided by the number of units ordered: $$P(x) = \frac{C(x)}{x} = \frac{5x^2 + 320}{x}.$$ 3. **Simplify the cost per unit function:** $$P(x) = \frac{5x^2}{x} + \frac{320}{x} = 5x + \frac{320}{x}.$$ 4. **Find the minimum of $P(x)$:** To minimize $P(x)$, take the derivative and set it to zero: $$P'(x) = 5 - \frac{320}{x^2}.$$ 5. **Set derivative equal to zero:** $$5 - \frac{320}{x^2} = 0.$$ 6. **Solve for $x^2$:** $$5 = \frac{320}{x^2} \implies 5x^2 = 320 \implies x^2 = \frac{320}{5} = 64.$$ 7. **Find $x$:** $$x = \sqrt{64} = 8.$$ 8. **Check the second derivative to confirm minimum:** $$P''(x) = \frac{640}{x^3} > 0 \text{ for } x > 0,$$ so $x=8$ is a minimum. **Final answer:** The cost per unit is minimized when ordering **8 units**. **Answer choice:** B An order of 8 units has a minimum cost per unit.