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 $$ x=10,000 $$.

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Alright, I need to find the marginal revenue at $$ x = 10,000 $$ for the revenue function $$ R(x) = 14x - 0.001x^2 $$. Let's break this down step by step.

First, I recall that marginal revenue is the derivative of the revenue function with respect to the number of units produced, $$ x $$. So, I need to find $$ R'(x) $$.

Given:
$$ R(x) = 14x - 0.001x^2 $$

To find the derivative, I'll differentiate each term separately.

1. The derivative of $$ 14x $$ with respect to $$ x $$ is $$ 14 $$.
2. The derivative of $$ -0.001x^2 $$ with respect to $$ x $$ is $$ -0.002x $$.

So, the marginal revenue function $$ R'(x) $$ is:
$$ R'(x) = 14 - 0.002x $$

Now, I need to find $$ R'(10,000) $$.

Plugging in $$ x = 10,000 $$:
$$ R'(10,000) = 14 - 0.002 \times 10,000 $$
$$ R'(10,000) = 14 - 20 $$
$$ R'(10,000) = -6 $$

Wait a second, a negative marginal revenue doesn't make sense in this context. Let me double-check my calculations.

Reviewing the derivative:
$$ R(x) = 14x - 0.001x^2 $$
$$ R'(x) = 14 - 0.002x $$

Calculating again:
$$ R'(10,000) = 14 - 0.002 \times 10,000 $$
$$ R'(10,000) = 14 - 20 $$
$$ R'(10,000) = -6 $$

Hmm, the math checks out. The negative value suggests that at $$ x = 10,000 $$ units, producing an additional unit would decrease revenue by $6,000. This could imply that the revenue function has reached its maximum at this point, and beyond this point, producing more units would start to reduce total revenue.

But let's consider the context. If the revenue is decreasing, perhaps the company has reached the point of diminishing returns. Maybe the cost of producing additional units is higher than the revenue generated, leading to a decrease in overall profit.

Alternatively, there might be an error in the revenue function. If the revenue is supposed to increase with more production, perhaps the coefficient of $$ x^2 $$ should be negative, making the function concave down. But as per the given function, it's currently concave up, which would mean that revenue increases initially but then starts to decrease after a certain point.

In conclusion, based on the provided revenue function $$ R(x) = 14x - 0.001x^2 $$, the marginal revenue at $$ x = 10,000 $$ is $-6,000. This indicates that producing an additional unit at this production level would result in a loss of $6,000 in revenue.

The marginal revenue at $$ x = 10,000 $$ is $-6,000.

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 \( x=10,000 \).

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