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08 Dec 2020, 04:00
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31% (02:54) correct
69%
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A Metro train from Mumbai to Gurgaon has capacity to board 900 people. The fare charged (in Rs) is defined by the function: \(f=(54−\frac{x}{32})^2\) where x is the number of the people per trip. How many people per trip will make the revenue for the trip maximum?
A. 1728
B. 592
C. 576
D. 484
E. 364
Are You Up For the Challenge: 700 Level Questions
A. 1728
B. 592
C. 576
D. 484
E. 364
Are You Up For the Challenge: 700 Level Questions
08 Dec 2020, 06:23
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Bunuel
Simplest although a bit calculation-intensive way would be to test the options.
Now total fare = \(f=(54−\frac{x}{32})^2*x\)
Eliminate A as 1728 is greater than the capacity of 900.
Substitute x as 592, then 576 and then 484 : you will find that 576 gives the highest amount.
No need to check for 364, as the values will taper on both sides of 576.
C
08 Dec 2020, 07:14
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CONCEPT: Functions and derivatives
At the point maximum, f'(x)=0
SOLUTION: Given, the number of people per trip is x.
The fare for each person =[(54−(x/32)]^2
The revenue from x people =x[(54−(x/32)]^2 =g(x)
where g(x)=[x(54−(x/32)]^2
When the revenue is the maximum, the marginal revenue equal to zero.
=>g′(x)=[54−(x/32)]^2−(2x/32)[(54-(x/32)]
For the maximum value of g(x):
g′(x)=0
⇒ 54−(x/32)=0 and (54−3x/32)=0
⇒ x=1728,x=576
Given the capacity of the train is 900, the number of people per trip = 576 (C)
Hope this helps you
Devmitra Sen
At the point maximum, f'(x)=0
SOLUTION: Given, the number of people per trip is x.
The fare for each person =[(54−(x/32)]^2
The revenue from x people =x[(54−(x/32)]^2 =g(x)
where g(x)=[x(54−(x/32)]^2
When the revenue is the maximum, the marginal revenue equal to zero.
=>g′(x)=[54−(x/32)]^2−(2x/32)[(54-(x/32)]
For the maximum value of g(x):
g′(x)=0
⇒ 54−(x/32)=0 and (54−3x/32)=0
⇒ x=1728,x=576
Given the capacity of the train is 900, the number of people per trip = 576 (C)
Hope this helps you
Devmitra Sen
