Question

watch the video and then solve the problem given below. click here to watch the video. the annual total revenue for a product is given by ( r(x)=20,000 x - 4 x^{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, units must be sold. (simplify your answer.)

Explanation:

Step1: Find the derivative of the revenue function

The revenue function is \( R(x) = 20000x - 4x^{2} \).
Using the power rule \((x^n)^\prime=nx^{n - 1}\), the derivative \(R^\prime(x)\) is:
\(R^\prime(x)=\frac{d}{dx}(20000x)-\frac{d}{dx}(4x^{2})\)
\(R^\prime(x)=20000-8x\)

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

To find the critical points, set \(R^\prime(x) = 0\).
\(20000-8x=0\)
Add \(8x\) to both sides: \(20000 = 8x\)
Divide both sides by \(8\): \(x=\frac{20000}{8}=2500\)

Step3: Check the second - derivative

The second - derivative \(R^{\prime\prime}(x)=\frac{d}{dx}(20000 - 8x)=-8\)
Since \(R^{\prime\prime}(x)=-8<0\), the function \(R(x)\) has a maximum at \(x = 2500\)

Answer:

\(2500\)