A manufacturer can sell x items of a commodity at price Rs (330-x) each and the cost of producing x items is Rs `(x^(2)+10x+12)`. Find MAR.

Answer

To find the maximum profit (MAR) for the manufacturer, we will follow these steps:

### Step 1: Define Selling Price and Cost Price
The selling price (SP) of one item is given as:
\[ SP = 330 - x \]
Thus, the selling price of \( x \) items is:
\[ \text{Total SP} = x \cdot (330 - x) = 330x - x^2 \]

The cost of producing \( x \) items is given as:
\[ \text{Cost Price (CP)} = x^2 + 10x + 12 \]

### Step 2: Calculate Profit
Profit (P) is defined as the difference between total selling price and total cost price:
\[ P = \text{Total SP} - \text{CP} \]
Substituting the values we have:
\[ P = (330x - x^2) - (x^2 + 10x + 12) \]
Simplifying this:
\[ P = 330x - x^2 - x^2 - 10x - 12 \]
\[ P = 330x - 2x^2 - 10x - 12 \]
\[ P = 320x - 2x^2 - 12 \]

### Step 3: Find the Maximum Profit
To find the maximum profit, we need to differentiate the profit function with respect to \( x \) and set the derivative equal to zero:
\[ \frac{dP}{dx} = 320 - 4x \]
Setting the derivative equal to zero for maximization:
\[ 320 - 4x = 0 \]
Solving for \( x \):
\[ 4x = 320 \]
\[ x = 80 \]

### Step 4: Verify Maximum Profit
To confirm that this value of \( x \) gives a maximum profit, we will check the second derivative:
\[ \frac{d^2P}{dx^2} = -4 \]
Since the second derivative is negative, this indicates that the profit function is concave down at \( x = 80 \), confirming a maximum.

### Conclusion
The maximum profit occurs when \( x = 80 \).

---