The demand function of an output is `x=106-2p`, where x is the output and p is the price per unit and average cost per unit is `5 + (x)/(50)`. Determine the number of units for maximum profit

Answer

To determine the number of units for maximum profit based on the given demand function and average cost function, we can follow these steps:

### Step 1: Understand the Demand Function
The demand function is given as:
\[ x = 106 - 2p \]
This implies that the output \( x \) is dependent on the price \( p \).

### Step 2: Express Price in Terms of Output
Rearranging the demand function to express \( p \) in terms of \( x \):
\[ 2p = 106 - x \]
\[ p = \frac{106 - x}{2} \]
This is our first equation.

### Step 3: Calculate Revenue
Revenue \( R \) is calculated as:
\[ R = p \cdot x \]
Substituting the expression for \( p \):
\[ R = \left(\frac{106 - x}{2}\right) \cdot x \]
\[ R = \frac{106x - x^2}{2} \]
This is our second equation.

### Step 4: Understand the Average Cost Function
The average cost per unit is given as:
\[ AC = 5 + \frac{x}{50} \]
To find the total cost \( C \), we multiply the average cost by the number of units:
\[ C = AC \cdot x = \left(5 + \frac{x}{50}\right) \cdot x \]
\[ C = 5x + \frac{x^2}{50} \]
This is our third equation.

### Step 5: Calculate Profit
Profit \( P \) is calculated as:
\[ P = R - C \]
Substituting the expressions for \( R \) and \( C \):
\[ P = \left(\frac{106x - x^2}{2}\right) - \left(5x + \frac{x^2}{50}\right) \]

### Step 6: Simplify the Profit Function
To simplify, we need a common denominator:
\[ P = \frac{106x - x^2}{2} - \left(5x + \frac{x^2}{50}\right) \]
Convert \( 5x \) to have a denominator of 50:
\[ 5x = \frac{250x}{50} \]
Now we can rewrite the profit function:
\[ P = \frac{106x - x^2}{2} - \left(\frac{250x}{50} + \frac{x^2}{50}\right) \]
\[ P = \frac{106x - x^2}{2} - \frac{250x + x^2}{50} \]

### Step 7: Combine the Terms
To combine these, we convert the first term to have a denominator of 50:
\[ P = \frac{2650x - 25x^2}{50} - \frac{250x + x^2}{50} \]
\[ P = \frac{2650x - 25x^2 - 250x - x^2}{50} \]
\[ P = \frac{2400x - 26x^2}{50} \]

### Step 8: Set the Derivative of Profit to Zero
To find the maximum profit, we take the derivative of \( P \) with respect to \( x \) and set it to zero:
\[ \frac{dP}{dx} = \frac{2400 - 52x}{50} = 0 \]
Solving for \( x \):
\[ 2400 - 52x = 0 \]
\[ 52x = 2400 \]
\[ x = \frac{2400}{52} = \frac{600}{13} \]

### Conclusion
The number of units for maximum profit is:
\[ x = \frac{600}{13} \]