Questions: Suppose we have a manager of a firm. The manager's assignment has provided cost estimates and the sales department has estimated demand for a new product. Analyze the situation and decide whether to produce the new product. The product has a production cost of 30 and a revenue function R(x) = 50x - 0.5x^2. The manager wants to maximize profit.

Suppose we have a manager of a firm. The manager's assignment has provided cost estimates and the sales department has estimated demand for a new product. Analyze the situation and decide whether to produce the new product. The product has a production cost of 30 and a revenue function R(x) = 50x - 0.5x^2. The manager wants to maximize profit.
Transcript text: Suppose we have a manager of a firm. The manager's assignment has provided cost estimates and the sales department has estimated demand for a new product. Analyze the situation and decide whether to produce the new product. The product has a production cost of $30 and a revenue function R(x) = 50x - 0.5x^2. The manager wants to maximize profit.
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Solution

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Solution Steps

To determine whether to produce the new product, we need to maximize the profit function. The profit function is given by the revenue function minus the cost function. The revenue function is \( R(x) = 50x - 0.5x^2 \) and the cost per unit is $30. The profit function is \( P(x) = R(x) - 30x \). We will find the derivative of the profit function, set it to zero to find critical points, and determine the maximum profit.

Step 1: Define the Profit Function

The profit function is given by the revenue function minus the cost function. The revenue function is \( R(x) = 50x - 0.5x^2 \) and the cost per unit is \( 30x \). Therefore, the profit function is: \[ P(x) = R(x) - 30x = 50x - 0.5x^2 - 30x = -0.5x^2 + 20x \]

Step 2: Find the Derivative of the Profit Function

To find the maximum profit, we need to find the derivative of the profit function and set it to zero: \[ P'(x) = \frac{d}{dx}(-0.5x^2 + 20x) = -x + 20 \]

Step 3: Solve for Critical Points

Set the derivative equal to zero to find the critical points: \[ -x + 20 = 0 \] \[ x = 20 \]

Step 4: Evaluate the Profit at Critical Points

Substitute the critical point back into the profit function to find the maximum profit: \[ P(20) = -0.5(20)^2 + 20(20) = -0.5(400) + 400 = -200 + 400 = 200 \]

Final Answer

The maximum profit is \(\boxed{200}\) when the production quantity is \( x = 20 \) units.

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