Comprehension
Directions for the following two questions:
Mr. David manufactures and sells a single product at a fixed price in a niche market. The selling price of each unit is Rs. 30. On the other hand, the cost, in rupees, of producing x units is \( 240 + bx + cx^2 \), where b and c are some constants. Mr. David noticed that doubling the daily production from 20 to 40 units increases the daily production cost by 66.67%. However, an increase in daily production from 40 to 60 units results in an increase of only 50% in the daily production cost. Assume that demand is unlimited and that Mr. David can sell as much as he can produce. His objective is to maximize the profit.
Question: 1
How many units should Mr. David produce daily?
How many units should Mr. David produce daily?
Show Hint
Use the given percentage increase in cost to set up equations for the cost parameters; then optimise profit via derivative.
Updated On: Jul 20, 2026
- 130
- 100
- 70
- 150
- Cannot be determined
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Given: Cost function \( C(x) = 240 + bx + cx^2 \).
From data:
When production increases from 20 to 40, cost increases by \( \frac{2}{3} \) of original cost at 20 units:
\[
C(40) - C(20) = \frac{2}{3} C(20).
\]
Similarly, from 40 to 60 units, cost increases by 50% of cost at 40 units:
\[
C(60) - C(40) = 0.5 \ C(40).
\]
Substitute \( C(x) \):
1st equation:
\[
[240 + 40b + 1600c] - [240 + 20b + 400c] = \frac{2}{3} [240 + 20b + 400c].
\]
Simplify:
\[
20b + 1200c = \frac{2}{3} (240 + 20b + 400c).
\]
2nd equation:
\[
[240 + 60b + 3600c] - [240 + 40b + 1600c] = 0.5 [240 + 40b + 1600c].
\]
Simplify:
\[
20b + 2000c = 0.5 (240 + 40b + 1600c).
\]
From first: Multiply through by 3: \( 60b + 3600c = 480 + 40b + 800c \) → \( 20b + 2800c = 480 \) → \( 5b + 700c = 120 \) → (i).
From second: \( 20b + 2000c = 120 + 20b + 800c \) → \( 1200c = 120 \) → \( c = 0.1 \).
From (i): \( 5b + 700(0.1) = 120 \) → \( 5b + 70 = 120 \) → \( 5b = 50 \) → \( b = 10 \).
Thus \( C(x) = 240 + 10x + 0.1x^2 \).
Revenue: \( R(x) = 30x \).
Profit: \( P(x) = 30x - (240 + 10x + 0.1x^2) = -0.1x^2 + 20x - 240 \).
Maximise profit: \( \frac{dP}{dx} = -0.2x + 20 = 0 \) → \( x = 100 \).
From second: \( 20b + 2000c = 120 + 20b + 800c \) → \( 1200c = 120 \) → \( c = 0.1 \).
From (i): \( 5b + 700(0.1) = 120 \) → \( 5b + 70 = 120 \) → \( 5b = 50 \) → \( b = 10 \).
Thus \( C(x) = 240 + 10x + 0.1x^2 \).
Revenue: \( R(x) = 30x \).
Profit: \( P(x) = 30x - (240 + 10x + 0.1x^2) = -0.1x^2 + 20x - 240 \).
Maximise profit: \( \frac{dP}{dx} = -0.2x + 20 = 0 \) → \( x = 100 \).
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Question: 2
What is the maximum daily profit, in rupees, that Mr. David can realize from his business?
What is the maximum daily profit, in rupees, that Mr. David can realize from his business?
Show Hint
Use the two percentage increases to find \(b\) and \(c\) in \(C(x)=240+bx+cx^2\), then maximise \(30x-C(x)\).
Updated On: Nov 4, 2026
- 620
- 760
- 840
- 920
- Cannot be determined
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Translate the percentage increases:
Doubling from 20 to 40 units raises the cost by 66.67% \(=\frac23\), so \(C(40)=\frac53C(20)\). Going from 40 to 60 units raises it by 50%, so \(C(60)=\frac32C(40)\).
Step 2: Find b and c:
From \(C(60)=1.5\,C(40)\): \(240+60b+3600c=360+60b+2400c\), so \(1200c=120\) and \(c=0.1\).
From \(3C(40)=5C(20)\): \(720+120b+4800c=1200+100b+2000c\), so \(20b+2800c=480\), i.e. \(b+140c=24\). With \(c=0.1\) this gives \(b=10\).
So \(C(x)=240+10x+0.1x^2\). (Check: \(C(20)=480,\ C(40)=800,\ C(60)=1200\), with ratios \(5/3\) and \(3/2\).)
Step 3: Profit function:
\(P(x)=30x-C(x)=20x-0.1x^2-240\). It is a downward parabola with maximum at \(x=\frac{20}{2(0.1)}=100\).
Step 4: Maximum profit:
\(P(100)=2000-1000-240=760\) rupees.
Final Answer:
The correct option is (2): 760.
Doubling from 20 to 40 units raises the cost by 66.67% \(=\frac23\), so \(C(40)=\frac53C(20)\). Going from 40 to 60 units raises it by 50%, so \(C(60)=\frac32C(40)\).
Step 2: Find b and c:
From \(C(60)=1.5\,C(40)\): \(240+60b+3600c=360+60b+2400c\), so \(1200c=120\) and \(c=0.1\).
From \(3C(40)=5C(20)\): \(720+120b+4800c=1200+100b+2000c\), so \(20b+2800c=480\), i.e. \(b+140c=24\). With \(c=0.1\) this gives \(b=10\).
So \(C(x)=240+10x+0.1x^2\). (Check: \(C(20)=480,\ C(40)=800,\ C(60)=1200\), with ratios \(5/3\) and \(3/2\).)
Step 3: Profit function:
\(P(x)=30x-C(x)=20x-0.1x^2-240\). It is a downward parabola with maximum at \(x=\frac{20}{2(0.1)}=100\).
Step 4: Maximum profit:
\(P(100)=2000-1000-240=760\) rupees.
Final Answer:
The correct option is (2): 760.
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