A bus company charges $15 per person for the first 50 passengers who buy tickets for a gold outing. The company offers a discount of $0.10 per ticket for each passenger in excess of 50. What number of passengers will provide maximum income for the bus company ?
Ehhhh... It's been a while since I've seen these questions.
@pooja195 @Luigi0210 @hartnn
I need help doing it , I'm a little bit stuck
what have u done so far
for first 50 passengers ticket cost will be $15 then from 51st passenget onwards the ticket cost will be $14.9
I need to find the number of passengers that will provide maximum income for the bus company
This is an optimization problem ,for calc 1
is the complete statement of the problem ?
I have I = # of passengers * price per passengers I = (50+x)(10-.10x) I = 500-5x+10x-0.1x^2 Do I take the derivative ? I'm stuck right there ,
That is the complete statement .
What do you mean by complete statement ?
I meant to write above I = (50+x)(15-0.1x)
yes - we need derivatives now, what is the derivative of this function?
dI/dx=10-0.2X
Sorry, my connection was acting fuzzy. So yes if you want to optimize you need to derive and make this equal to 0
0=10-0.2x ? 0.2x=10 X= 50 ?
Let me show you: dI/dx= 5-(0.1)(2)x
I = 500-5x+10x-0.1x^2 =5x-(0.1)(2)x
you see?
So this new one, could you equal it to 0 and find the x?
I made a mistake originally , I fixed it to be : I= (50+x)(15-0.10x) Which equals to I = 750+10x-0.1x^2 I wrote (50+x)(10-0.10x) which was wrong . The price is $15 not $10
Oh, ok. So the derivative would be?
The derrative of that would be dI/dx= 10-0.2X
Yep! So when its equal to zero?
0=10-.2x .2x=10 X=50?
Yess!!!
so at 50 persons the price get to its maximum :)
After what do I do next ? Plug it in the original equation ?
Nope! That's your answer X = 50 !
Oh ok
Thank you so much!!!
You're most welcome! Good luck with your studies! :D
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