Question

Profit = Total Revenue -total cost . A maker of custom high-fashion boots has an annual total cost function in $ given by: C(x)=3600+100x+2x^2 and an annual total revenue function $ given by: R(x)=500x-2x^2. For both functions, x is the number of pairs of boots sold. Write the profit function. Then determine the number of pairs of boots that will maximize annual profit.

Answer 1

The number of pairs of boots that will maximize the profit is 80 pairs of boots

Explanation:

`\begin{gathered} \text{Given that} \\ \text{ Profit = Total revenue }-\text{ total cost} \\ \text{ Total cost function is given as} \\ C(x)=3600+100x+2x^2 \\ \text{ Revenue function is given as} \\ R(x)=500x-2x^2 \\ \text{Profit = 500x - 2x}^2-(3600+100x+2x^2) \\ \text{Profit = 500x -2x}^2-3600-100x-2x^2 \\ \text{Profit = -2x}^2-2x^2\text{ }+\text{ 500x - 100x - 3600} \\ \text{Profit = -4x}^2\text{ + 400x - 3600} \\ \text{Profit = -x}^2\text{ + 100x - 900} \\ \text{The profit function is given as -x}^2\text{ + 100x - 900} \end{gathered}`

Question 2:

Calculate the number of pairs of boots that will maximize annual profit

`\begin{gathered} \text{Given the profit function to be} \\ Profit=-x^2\text{ + 100x - 900} \\ \text{Let profit = 0} \\ -x^2\text{ + 100x - 900 =0} \\ \text{ Using the general quadratic formula} \\ x\text{ = }\frac{-b\pm\sqrt[]{b^2\text{ - 4ac}}}{2a} \\ \text{let a = -1, b= 100 and c = -900} \\ x\text{ = }\frac{-100\pm\sqrt[]{100^2\text{ - 4 }\cdot\text{ (-1) (-900)}}}{-2\text{ x1}} \\ x\text{ = }\frac{-100\pm\sqrt[]{10000\text{ - 3600}}}{-2} \\ x\text{ = }\frac{-100\pm\sqrt[]{6400}}{-2} \\ x\text{ = }\frac{-100\text{ }\pm80}{-2} \\ x\text{ = }\frac{-100\text{ + 80}}{-2}\text{ or }\frac{-100\text{ - 80}}{-2} \\ x\text{ = }(-20)/(-2)\text{ or }(-160)/(-2) \\ x\text{ = 10 or 80} \\ \text{Hence, the number of pairs of boots that will max}imize\text{ the profit is 80 pairs of boots} \end{gathered}`