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Updated on: 17 May 2021, 14:17
Originally posted by sujoykrdatta on 14 May 2021, 10:58.
Last edited by sujoykrdatta on 17 May 2021, 14:17, edited 1 time in total.
Last edited by sujoykrdatta on 17 May 2021, 14:17, edited 1 time in total.
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C
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Difficulty:
35%
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Question Stats:
74% (02:10) correct
26%
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There is a certain cable operator who charges Rs 300 per subscriber. He presently has 400 subscribers. For every Rupee he increases the subscription fees by, he would lose 1 subscriber. What should be the new subscription charge so that he maximizes his total revenue?
(Rs = Rupees: the currency used in India)
A) Rs 310
B) Rs 325
C) Rs 350
D) Rs 380
E) Rs 420
(Rs = Rupees: the currency used in India)
A) Rs 310
B) Rs 325
C) Rs 350
D) Rs 380
E) Rs 420
14 May 2021, 12:21
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sujoykrdatta
Operators initial income I=300×400
If he increase 'x' Rs he will lose 'x' subscriber.
So, I'= (300+x) (400-x)
To find maximum income we have to differentiate this.
\(\frac{d(i)}{dx}\)=(300+x) (-1) + (400-x) (1)
= 400- x- 300- x
=100- 2x
So, 100- 2x= 0 (For maximum and minimum d(I)/dx=0)
=>x= 50
\(\\
\frac{d^2(I)}{dx}\)=0-2=-2
-2<0 So, x=50 is maximum.
The answer is C
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17 May 2021, 14:16
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sujoykrdatta
Revenue = Fees per subscriber * Number of subscribers
For every Rupee he increases the subscription fees by, he would lose 1 subscriber
Let us assume that he increases the fees by Rs x; hence, the number of subscribers would reduce by x
Thus, revenue = Rs (300 + x)(400 - x)
Observe that we need to maximize the product of the terms (300+x) and (400-x)
We know that if the sum of 2 terms is a constant, their product is maximized when the terms are equal (results from AM - GM inequality)
For example: If a+b = 10, the maximum value of ab = 5*5 = 25
Similarly, here, the sum of the terms (300+x) and (400-x) is 700
Hence, the product will be maximized when the terms are 350 each
Note: We thus have: 300+x = 350 => x = 50
New fees per subscriber = Rs 350
Answer C
