Given that
The cost for producing items is 40x + 300 and the revenue collected is 80x - 0.5x^2
And we have to write an equation for the profit and also we have to find the profit.
Explanation -
The profit is basically the difference between revenue and cost.
So the profit will be
![Profit=Revenue-cost](https://img.qammunity.org/2023/formulas/mathematics/college/unyowx4fz22thxgxe4hgv0k95pggotfzuq.png)
Substituting the values
![\begin{gathered} Profit=80x-0.5x^2-(40x+300) \\ Profit=80x-0.5x^2-40x-300 \\ P=40x-0.5x^2-300 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/un053yssoqvz17laonm7q3yefm89lpt1lj.png)
(a). The profit equation will be, P = 40x - 0.5x^2 - 300
(b). Here the profit given is $300 (P=300) and we have to find the values of x.
Then,
![\begin{gathered} 300=40x-0.5x^2-300 \\ 600=40x-0.5x^2 \\ 0.5x^2-40x+600=0 \\ Dividing\text{ by 0.5 we have} \\ x^2-80x+1200=0 \\ x^2-60x-20x+1200=0 \\ x(x-60)-20(x-60)=0 \\ (x-20)(x-60)=0 \\ x=20\text{ and x = 60} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/fb34x1h2o39lie3a4no5gh71ipiyqb3dch.png)
So the two values of x are 20 and 60.
(c). Now we have to check whether the profit can be $15000 or not.
For this, we have to put P = 15000 in the quadratic equation and if we get a solution for x then it is possible otherwise it is not.
So, substituting P = 15000
![\begin{gathered} 15000=40x-0.5x^2-300 \\ 0.5x^2-40x+15300=0 \\ Dividing\text{ by 0.5 in the above equation} \\ x^2-80x+30600=0 \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/y2yx908z5prup6a8v8plfmrbjqd92ao7es.png)
The solution to this equation is not possible. So Company cannot make a profit of $15000.
Hence the final answers are(a). P = 40x - 0.5x^2 - 300(b). 20 and 60(c). No