Total profit P is the difference between total revenue R and total cost C. Given the following​ total-revenue and​ total-cost functions, find the total​ profit, the maximum value of the total​ profit, and the value of x at which it occurs.

R(x)=1300x-x^2
C(x)=3300+20x

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Step-by-step explanation: x=495

,P=241,525

Step-by-step explanation:

Given that

Profit=Total revenue - total cost

P= R -C

Also given that

So

Above equation is the total profit in terms if x.

Now to find maximum value of P we have to differentiate above equation with respect to x

So

⇒x=495

So total profit at x=445  ,P=241,525

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]

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