A manufacturer can sell 'x' items at a price of Rs `(250-x)` each. The cost of producing 'x' items is Rs `(x^(2)-50x+12)`. Determine the number of items to be sold so that he can make maximum profit.

Answer

To determine the number of items to be sold for maximum profit, we will follow these steps:

### Step 1: Define the Selling Price and Cost
The selling price per item is given by:
\[ \text{Selling Price} = 250 - x \]
Thus, the total selling price for \(x\) items is:
\[ \text{Total Selling Price} = x(250 - x) = 250x - x^2 \]

The cost of producing \(x\) items is given by:
\[ \text{Cost} = x^2 - 50x + 12 \]

### Step 2: Define the Profit Function
Profit is defined as the total selling price minus the total cost:
\[ \text{Profit} = \text{Total Selling Price} - \text{Cost} \]
Substituting the expressions we found:
\[ \text{Profit} = (250x - x^2) - (x^2 - 50x + 12) \]
Simplifying this:
\[ \text{Profit} = 250x - x^2 - x^2 + 50x - 12 \]
\[ \text{Profit} = -2x^2 + 300x - 12 \]

### Step 3: Differentiate the Profit Function
To find the maximum profit, we need to differentiate the profit function with respect to \(x\):
\[ \frac{dP}{dx} = -4x + 300 \]

### Step 4: Set the Derivative to Zero
To find the critical points, set the derivative equal to zero:
\[ -4x + 300 = 0 \]
Solving for \(x\):
\[ 4x = 300 \implies x = 75 \]

### Step 5: Verify Maximum Profit
To confirm that this critical point is a maximum, we will take the second derivative of the profit function:
\[ \frac{d^2P}{dx^2} = -4 \]
Since the second derivative is negative, this indicates that the profit function is concave down, confirming that \(x = 75\) gives a maximum profit.

### Conclusion
The number of items that should be sold to maximize profit is:
\[ \boxed{75} \]