Given a function as shown below:
[tex]P(x)=550x-1240-2x^2[/tex][tex]\begin{gathered} for\text{ profit maximization } \\ P(x)=550x-1240-2x^2 \\ \frac{dp(x)}{dx}=550-4x \\ 0\text{ = 550 -4x} \\ 4x\text{ = 550} \\ x=\text{ }\frac{550}{4} \\ x=137.5 \\ x=\text{ 138 (nearest whole number)} \end{gathered}[/tex]
Hence to maximize the profit at least 138 items must be sold
And the maximum profit that can be earned is derived by substituting 138 it to p(x)
[tex]\begin{gathered} P(x)=550x-1240-2x^2 \\ P(x)=550(138)-1240-2(138)^2 \\ P(x)=\text{ 36572}.00 \end{gathered}[/tex]
Hence the maximum profit is $36572.00
The minimum item that must be sold to make a profit can be found by solving the following quadratic inequality:
[tex]P(x)=550x-1240-2x^2\ge0[/tex]
And the solution is given as
[tex]\begin{gathered} \frac{-\sqrt{73145}+275}{2}\le\: x\le\frac{\sqrt{73145}+275}{2} \\ 2.27333\le\: x\le\: 272.72666 \end{gathered}[/tex]
Hence, the company must sell a minimum of 3 items to make a profit
