**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)

## 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−x32)2f=(54−x32)2 where x is the number of the people per trip. How many people per trip will make the revenue Print

Modified on: Tue, 12 Oct, 2021 at 1:04 PM

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