Question 7 A company’s profit P(x)P(x) (in dollars) depends on the number of items xx produced per week: P(x)=−5x2+300x−2000P(x) = -5x^2 + 300x - 2000 (a) Determine the production level xx that maximizes profit. (b) Compute the maximum profit. (c) Find the production levels xx at which the profit is zero. Show solutionHide solution+Question 7 - Solution (a) Maximize Profit: P′(x)=ddx(−5x2+300x−2000)=−10x+300P'(x) = \frac{d}{dx}(-5x^2 + 300x - 2000) = -10x + 300 Set P′(x)=0P'(x) = 0: −10x+300=0⇒x=30-10x + 300 = 0 \quad \Rightarrow \quad x = 30 Second derivative: P″(x)=−10<0P''(x) = -10 < 0 Hence, profit is maximized at x=30x = 30. (b) Maximum Profit: P(30)=−5(30)2+300(30)−2000=−4500+9000−2000=2500P(30) = -5(30)^2 + 300(30) - 2000 = -4500 + 9000 - 2000 = 2500 (c) Break-even Points (Profit = 0): −5x2+300x−2000=0-5x^2 + 300x - 2000 = 0 Divide through by 5: x2−60x+400=0x^2 - 60x + 400 = 0 Solve using the quadratic formula: x=60±3600−16002=60±20002=30±105x = \frac{60 \pm \sqrt{3600 - 1600}}{2} = \frac{60 \pm \sqrt{2000}}{2} = 30 \pm 10\sqrt{5} Maximum profit occurs at x=30 units, Pmax=2500,Profit is zero at x=30−105,30+105.\boxed{ \begin{aligned} &\text{Maximum profit occurs at } x = 30 \text{ units, } P_{\max} = 2500, \\ &\text{Profit is zero at } x = 30 - 10\sqrt{5}, \; 30 + 10\sqrt{5}. \end{aligned} }
