Question

the annual total revenue for a product is given by ( r(x) = 63,000x - 7x^{2} ) dollars, where ( x ) is the number of units sold. to maximize revenue, how many units must be sold? what is the maximum possible annual revenue?
to maximize revenue, ( square ) units must be sold.
(simplify your answer.)

Explanation:

Step1: Find the derivative of the revenue function

The revenue function is \( R(x) = 63000x-7x^{2}\).
Using the power rule \((x^n)^\prime=nx^{n - 1}\), the derivative \(R^\prime(x)\) is:
\(R^\prime(x)=\frac{d}{dx}(63000x)-\frac{d}{dx}(7x^{2})\)
\(R^\prime(x)=63000-14x\)

Step2: Set the derivative equal to zero and solve for \(x\)

To find the critical points, set \(R^\prime(x) = 0\).
\(63000-14x=0\)
Add \(14x\) to both sides: \(14x = 63000\)
Divide both sides by \(14\): \(x=\frac{63000}{14}=4500\)

Step3: Check the second - derivative

The second - derivative \(R^{\prime\prime}(x)=\frac{d}{dx}(63000 - 14x)=-14\)
Since \(R^{\prime\prime}(x)=-14<0\), the function \(R(x)\) has a maximum at \(x = 4500\)

Answer:

\(4500\)