Questions: Suppose a company has its total cost for a product given by C(x)=200 x+10,000 dollars and its total revenue given by R(x)=250 x-0.01 x^2 dollars, where x is the number of units produced and sold. Find the marginal revenue function R'(x)=250-0.02 x Find " R(72)-R(71) " and " R' (72) " and tell what each represents or predicts. Find the profit function and marginal profit function. P'(x) 50-0.02 x At which point does the profit reach its maximum? and the maximum profit? 52500

Suppose a company has its total cost for a product given by C(x)=200 x+10,000 dollars and its total revenue given by R(x)=250 x-0.01 x^2 dollars, where x is the number of units produced and sold.
Find the marginal revenue function
R'(x)=250-0.02 x
Find " R(72)-R(71) " and " R' (72) " and tell what each represents or predicts.
Find the profit function and marginal profit function.
P'(x)  50-0.02 x
At which point does the profit reach its maximum? and the maximum profit?
52500
Transcript text: Suppose a company has its total cost for a product given by $C(x)=200 x+10,000$ dollars and its total revenue given by $R(x)=250 x-0.01 x^{2}$ dollars, where x is the number of units produced and sold. Find the marginal revenue function \[ R^{\prime}(x)=250-0.02 x \] Find " $R(72)-R(71)$ " and " $R$ ' $(72)$ " and tell what each represents or predicts. Find the profit function and marginal profit function. \[ P^{\prime}(x) \quad 50-0.02 x \] At which point dose the profit reach its maximum? and the maximum profit? 52500
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Solution

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Solution Steps

Solution Approach

10-1. To find the marginal revenue function, we need to take the derivative of the revenue function \( R(x) = 250x - 0.01x^2 \) with respect to \( x \).

10-2. To find \( R(72) - R(71) \), we calculate the revenue at 72 units and subtract the revenue at 71 units. \( R'(72) \) is the derivative of the revenue function evaluated at \( x = 72 \), which represents the approximate change in revenue for an additional unit sold at 72 units.

10-3. The profit function \( P(x) \) is the difference between the revenue function \( R(x) \) and the cost function \( C(x) \). The marginal profit function is the derivative of the profit function with respect to \( x \).

Step 1: Marginal Revenue Function

To find the marginal revenue function, we differentiate the revenue function \( R(x) = -0.01x^2 + 250x \) with respect to \( x \): \[ R'(x) = 250 - 0.02x \]

Step 2: Calculate \( R(72) - R(71) \) and \( R'(72) \)

Next, we compute the revenue at \( x = 72 \) and \( x = 71 \): \[ R(72) \approx 17948.16, \quad R(71) \approx 17699.59 \] Thus, the difference in revenue is: \[ R(72) - R(71) \approx 248.57 \] We also find the marginal revenue at \( x = 72 \): \[ R'(72) \approx 248.56 \]

Step 3: Profit Function and Marginal Profit Function

The profit function \( P(x) \) is given by the difference between revenue and cost: \[ P(x) = R(x) - C(x) = (-0.01x^2 + 250x) - (200x + 10000) = -0.01x^2 + 50x - 10000 \] The marginal profit function is the derivative of the profit function: \[ P'(x) = 50 - 0.02x \]

Final Answer

  • Marginal Revenue Function: \( R'(x) = 250 - 0.02x \)
  • \( R(72) - R(71) \approx 248.57 \)
  • \( R'(72) \approx 248.56 \)
  • Profit Function: \( P(x) = -0.01x^2 + 50x - 10000 \)
  • Marginal Profit Function: \( P'(x) = 50 - 0.02x \)

\[ \boxed{R'(x) = 250 - 0.02x, \quad R(72) - R(71) \approx 248.57, \quad R'(72) \approx 248.56} \]

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