# Find maximum profit

The cost of producing appliances is $$C(x) = 300 + 2.2x^2$$ where $$x$$ is the number produced per week.

If they are sold for $110 each, how many should be produced for maximum profit? Solution: 25 Approach: Profit = Revenue - Cost $$P(x) = R(x) - C(x)$$ Let the cost of producing x number of units per week be $$C(x) = 300 + 2.2x^2$$ and the revenue $$R(x) = 110x$$ for selling x number of units. Hence profit is $$P(x) = 110x - 300 - 2.2x^2$$. To find the maximum profit, the derivative of $$P(x) = 0$$. $$P'(x) = 110 - 2.2x$$ $$110 - 2.2x = 0$$ $$x = 50$$ The method above gives me x = 50 where the solution says 25. Is my understanding and approach correct? ## 2 Answers In spirit of the answer given by @trula, to double check your answer you may find the $$x$$ coordinate of vertex: $$P(x) = 110x - 300 - 2.2x^2$$ $$x = -\dfrac{b}{2a}=-\dfrac{110}{2\times 2.2} = 25$$ • Thank you! Where did you get the$x = -\dfrac{b}{2a}=-\dfrac{110}{2\times 2.2}\$ formula? – Moonshine45 Nov 9 '19 at 5:37
• The Vertex of a parabola y=ax^2+bx+c is at -b/2a since ax^2+bx+c =a(x^2+b/a*x+(b/2a)^2-(b/2a)^2)+c=a*(x-b/2a)^2-b^2/2a+c – trula Nov 9 '19 at 5:50
• Yeah but please use the vertex formula only to double check. Because what you're using with calculus is the general method. Your calculus method for any function P(x). For more on finding vertex of quadratic without calculus see this – AgentS Nov 9 '19 at 6:05

your derivative is wrong. 𝑃′(𝑥)=110−4.4𝑥

• thanks! bit of an oversight there – Moonshine45 Nov 9 '19 at 5:36