Answer:
63568 dollars
Explanation:
Given that in your computer store you charge $1012 per computer sold. Your costs are given by the equation
![C(x) = 4x^2 + 4x - 80](https://img.qammunity.org/2020/formulas/mathematics/college/mqtj02yl9bxl3mfclszkmmi4jkth3p2857.png)
where x is the number of computers sold.
Revenue = sales price *no of computers sold
= 1012x
Profit = Revenue - cost
=
![1012x-( 4x^2 + 4x - 80)\\= 1008x-4x^2+80\\](https://img.qammunity.org/2020/formulas/mathematics/college/vhn4llnwngrmvoidjam33usk4px6sqr8fn.png)
Use derivative test to find maximum profit
![C'(x) = 1008-8x \\C''(x)= -x<0](https://img.qammunity.org/2020/formulas/mathematics/college/171m5y14f7uv76tsakxkjwlzueqy4ytfd9.png)
Equate I derivative to 0
x= 128
i.e. if 128 computers are manufactured and sold profit wouldbemaximum
Profit maximum=
![1008(128)-4(128)^2+80\\=63568](https://img.qammunity.org/2020/formulas/mathematics/college/u903h88kq8mxosiml2oj14n2x59fwhwvx2.png)