Manufacturer can sell x items at a price of rupees `(5-x/(100))`each. The cost price of x items is Rs `(x/5+500)`. Find the number of items he should sell to earn maximum profit

Options

A. `x=140`
B. `x=240`
C. `x=340`
D. `x=440`

Explanation

To solve the problem step by step, we will first define the selling price, cost price, and profit functions based on the information provided. Then, we will find the number of items that should be sold to maximize profit.

### Step 1: Define the Selling Price
The selling price per item is given by:
\[ P(x) = 5 - \frac{x}{100} \]
Thus, the total selling price for \(x\) items is:
\[ \text{Total Selling Price} = x \cdot P(x) = x \left(5 - \frac{x}{100}\right) = 5x - \frac{x^2}{100} \]

### Step 2: Define the Cost Price
The cost price for \(x\) items is given by:
\[ \text{Cost Price} = \frac{x}{5} + 500 \]

### Step 3: Define the Profit Function
Profit is defined as the total selling price minus the cost price:
\[ \text{Profit} = \text{Total Selling Price} - \text{Cost Price} \]
Substituting the expressions we derived:
\[ \text{Profit} = \left(5x - \frac{x^2}{100}\right) - \left(\frac{x}{5} + 500\right) \]
Simplifying this, we get:
\[ \text{Profit} = 5x - \frac{x^2}{100} - \frac{x}{5} - 500 \]
To combine the terms, we can express \(5x\) as \(\frac{25x}{5}\):
\[ \text{Profit} = \frac{25x}{5} - \frac{x}{5} - \frac{x^2}{100} - 500 = \frac{24x}{5} - \frac{x^2}{100} - 500 \]

### Step 4: Differentiate the Profit Function
To find the maximum profit, we need to differentiate the profit function with respect to \(x\):
\[ \frac{d(\text{Profit})}{dx} = \frac{d}{dx}\left(\frac{24x}{5} - \frac{x^2}{100} - 500\right) \]
Calculating the derivative:
\[ \frac{d(\text{Profit})}{dx} = \frac{24}{5} - \frac{2x}{100} = \frac{24}{5} - \frac{x}{50} \]

### Step 5: Set the Derivative to Zero
To find the critical points, set the derivative equal to zero:
\[ \frac{24}{5} - \frac{x}{50} = 0 \]
Solving for \(x\):
\[ \frac{x}{50} = \frac{24}{5} \implies x = 50 \cdot \frac{24}{5} = 240 \]

### Step 6: Verify Maximum Profit
To confirm that this critical point is a maximum, we will find the second derivative:
\[ \frac{d^2(\text{Profit})}{dx^2} = -\frac{2}{100} = -\frac{1}{50} \]
Since the second derivative is negative, this indicates that the profit function has a maximum at \(x = 240\).

### Conclusion
The manufacturer should sell **240 items** to earn the maximum profit.