We have to find the expression for the profit in function of the number of units x.
We have an expression for the revenue R(x) and for the cost C(x):
![\begin{gathered} R(x)=5x^2+2x-80 \\ C(x)=5x^2-x+100 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/x01jgzw1t45blcxa4bjzhclf7icso0k0o4.png)
If the profit, that we will call P(x), is the difference between the revenue and the cost, we can calculate it like this:
![\begin{gathered} P(x)=R(x)-C(x) \\ P(x)=(5x^2+2x-80)-(5x^2-x+100) \\ P(x)=5x^2+2x-80-5x^2+x-100 \\ P(x)=(5x^2-5x^2)+(2x+x)+(-80-100) \\ P(x)=3x-180 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jyf7m49g8ntozh6b6vg9oiy4sr06elnd8g.png)
We can now calculate the profit for the sale of 1000 videogames (x = 1000) as:
![P(1000)=3\cdot1000-180=3000-180=2820](https://img.qammunity.org/2023/formulas/mathematics/college/pwg1rbrlaadsbcuaezb0jn1ocpf0dxc71v.png)
Answer:
a) P(x) = 3x - 180
b) The profit is $2820