Question
he revenue (in thousands of dollars) from producing \( x \) units of an item is nodeled by \( R(x)=14 x-0.001 x^{2} \). b. Find the marginal revenue at \( x=10,000 \).
Ask by Kelley Pritchard. in the United States
Jan 24,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The marginal revenue at \( x = 10,000 \) is -6.
Solution
It seems there was an issue with the calculation. Let's try to find the derivative of the revenue function \( R(x) = 14x - 0.001x^2 \) with respect to \( x \) manually.
The derivative of \( R(x) \) with respect to \( x \) is given by:
\[ R'(x) = 14 - 0.002x \]
Now, we need to find the marginal revenue at \( x = 10,000 \) by substituting \( x = 10,000 \) into the derivative \( R'(x) \).
Evaluate the expression by following steps:
- step0: Evaluate:
\(14-0.002x\)
- step1: Substitute:
\(14-0.002\times 10000\)
- step2: Multiply the numbers:
\(14-20\)
- step3: Subtract the numbers:
\(-6\)
The marginal revenue at \( x = 10,000 \) is -6.
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Simplify this solution Bonus Knowledge
To find the marginal revenue at \( x = 10,000 \), we first need to calculate the derivative of the revenue function \( R(x) = 14x - 0.001x^2 \). The derivative, which represents the marginal revenue, is found as follows: \[ R'(x) = 14 - 0.002x \] Now, we can find the marginal revenue when \( x = 10,000 \): \[ R'(10,000) = 14 - 0.002 \times 10,000 = 14 - 20 = -6 \] Thus, the marginal revenue at \( x = 10,000 \) is \(-6\) thousand dollars, indicating that producing an additional unit would reduce revenue by $6,000. This means that you might want to reconsider scaling production at this level!
