Answer:
The profit function is:
[tex]P(x)=-x^2+1280x-3300[/tex]
The maximum value is 406, 300 occurring when x = 640.
Step-by-step explanation:
The revenue function is:
[tex]R(x)=1300x-x^2[/tex]
And the cost function is:
[tex]C(x)=3300+20x[/tex]
Then the total profit function will be:
[tex]P(x)=R(x)-C(x)=(1300x-x^2)-(3300+20x)=-x^2+1280x-3300[/tex]
This is a quadratic function.
Therefore, the maximum value of the total profit will occur at its vertex point.
The vertex of a quadratic is given by:
[tex]\displaystyle \Big(-\frac{b}{2a}, f\Big(-\frac{b}{2a}\Big)\Big)[/tex]
In this case, a = -1, b = 1280, and c = -3300.
Then the point at which the maximum profit occurs is at:
[tex]\displaystyle x=-\frac{1280}{2(-1)}=640[/tex]
And the maximum profit will be:
[tex]P(640)=-(640)^2+1280(640)-3300=406300[/tex]
