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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