Question

A manufacturer sells $x$ items at a price of ₹$(6 - \frac{x}{40})$ each. The cost price of $x$ items is ₹$(\frac{x}{5} + 193)$. The maximum profit in ₹ is ______.

• A. 134.4
• B. 144.3
• C. 143.4
• D. 133.4

Hint:

Be careful! The problem states the selling price is per item, meaning you must multiply by $x$ to get total revenue. However, the cost is given for "$x$ items", meaning it is already the total cost. Don't multiply the cost by $x$ again!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Profit $P(x) = \text{Total Revenue} - \text{Total Cost}$. Revenue is $\text{Price} \times x$.

Step 2: Formula Application:
$R(x) = x(6 - \frac{x}{40}) = 6x - \frac{x^2}{40}$. $P(x) = (6x - \frac{x^2}{40}) - (\frac{x}{5} + 193) = \frac{29}{5}x - \frac{x^2}{40} - 193$.

Step 3: Explanation:
To maximize profit, $P'(x) = 0$. $P'(x) = \frac{29}{5} - \frac{2x}{40} = 0 \implies \frac{x}{20} = \frac{29}{5} \implies x = 116$. Substitute $x = 116$ into $P(x)$: $P(116) = \frac{29 \times 116}{5} - \frac{116^2}{40} - 193 = 672.8 - 336.4 - 193 = 143.4$.

Step 4: Final Answer:
The maximum profit is ₹143.4.