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 Question:
This is an economic optimization problem. We are given the selling price per item and the total cost for $x$ items. We must formulate a Profit function and use calculus to maximize it.

Step 2: Detailed Explanation:
1. Define the Total Revenue Function $R(x)$:
Revenue = (Number of items) $\times$ (Selling price per item)
$R(x) = x \left(6 - \frac{x}{40}\right) = 6x - \frac{x^2}{40}$
2. Define the Total Cost Function $C(x)$:
This is explicitly given:
$C(x) = \frac{x}{5} + 193$
3. Define the Profit Function $P(x)$:
Profit = Total Revenue - Total Cost
$P(x) = R(x) - C(x)$
$P(x) = \left(6x - \frac{x^2}{40}\right) - \left(\frac{x}{5} + 193\right)$
$P(x) = -\frac{x^2}{40} + 6x - \frac{x}{5} - 193$
$P(x) = -\frac{x^2}{40} + \frac{30x - x}{5} - 193$
$P(x) = -\frac{x^2}{40} + \frac{29x}{5} - 193$
4. Maximize the Profit Function:
Find the first derivative $P'(x)$ and set it to zero to find the critical point:
$P'(x) = -\frac{2x}{40} + \frac{29}{5} = -\frac{x}{20} + \frac{29}{5}$
Set $P'(x) = 0$:
$\frac{x}{20} = \frac{29}{5}$
$x = 20 \times \frac{29}{5} = 4 \times 29 = 116$
So, maximum profit occurs when 116 items are sold.
5. Calculate the Maximum Profit:
Substitute $x = 116$ back into the original Profit function $P(x)$:
$P(116) = -\frac{(116)^2}{40} + \frac{29(116)}{5} - 193$
Calculate the terms:
$\frac{(116)^2}{40} = \frac{13456}{40} = 336.4$
$\frac{29 \times 116}{5} = \frac{3364}{5} = 672.8$
$P(116) = -336.4 + 672.8 - 193$
$P(116) = 336.4 - 193 = 143.4$

Step 3: Final Answer:
The maximum profit is ₹143.4, matching option (c).