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Thread starter vedmachka Start date Oct 16, Joined Oct 15, Messages 6. This is due tomorrow and I am at a loss. I know A Bus Company Has 4000 Passengers to solve these problems, but this one just got me stuck for some reason. Any help would be great. What fare should the company charge to maximize their revenue? Any help would Gaslight Company greatly appreciated I don't know how to solve it other than what I already did Joined Oct 6, Messages 10, Re: Easy problem, but got me A Bus Company Has 4000 Passengers lost Hi there: The revenue equation is the number of passengers times the fare.

It looks like you already wrote it but set it equal to zero? If x represents the number of cent fare increases, then the daily revenue is a function of x. The maximum revenue is given by the y-coordinate at the vertex of the parabola.

But, this exercise does not ask for the maximum revenue. You can find the x-coordinate of the vertex two ways. It's halfway between the two roots, or use the formula. I'm not sure what you mean by "substitute a value into the dependent variable", since R x is the dependent variable. Denis Senior Member. Joined Feb 17, Messages 1, 1840 Brewing Company said:.

I restricted the increase to multiples of. Subhotosh Khan Super Moderator Staff member. Joined Jun 18, Messages 20, Ok, This is due tomorrow and I am at a loss. It could be that this is your problem. The equation above is the revenue function - and you want to maximize it - NOT set to 0. If you use derivative, then derivative of the function above is set to zero. However, my teacher is telling me that my method was NOT grade appropriate and that I should write this equation as a revenue function so that I Smart Company Alabama substitute a value into the dependent variable.

Thank you guys for your replies, however that is the solution that I got as well. I did the question using derivatives. The teacher said not to use derivatives. Still lost OR, you can use the quadratic formula to get the two roots Even though those solutions may be easier, and may also produce the correct answer, they are not in the grade 11 curriculum.

I can't seem to solve this any other way. This statement is clearly false, but I'm glad you're happy. You must log in or register to reply here.

## SOLUTION: A bus company has 4000 passengers daily, each ...

Question 123826: A bus company has 4000 passengers daily, each paying a fare of $2. For each $.015 increase, the company estimates that it will lose 40 passengers per day. For each $.015 increase, the company estimates that it will lose 40 passengers per day.…

## SOLUTION: A bus company has 4000 passengers daily, each ...

Question 384489: A bus company has 4000 passengers daily, each paying a fee of 2$. For each 0.15$ increase, the company estimates that it will lose 40 passengers. For each 0.15$ increase, the company estimates that it will lose 40 passengers.…

## SOLUTION: A bus company has 4000 passengers daily, each ...

Question 871669: A bus company has 4000 passengers daily, each paying a fare of $2. For each $0.15 increase, the company estimates that it will lose 40 passengers per day. For each $0.15 increase, the company estimates that it will lose 40 passengers per day.…

## SOLUTION: A bus company has 4000 passengers daily, each ...

Question 1049512: A bus company has 4000 passengers daily, each paying a fare of $2. For each $0.15 increase, the company estimates that it will lose 40 passengers per day. For each $0.15 increase, the company estimates that it will lose 40 passengers per day.…

## A bus company has 4000 passengers daily, each paying a a ...

May 19, 2018 · A bus company has 4000 passengers daily, each paying a fare of $2. For each $0.15 increase, the company estimates that it will lose 40 passengers. If the company needs to take in $10,450 per day to stay in business, what fare . asked by Emme on January 19, 2013; Math…

## a bus company has 4000 passengers daily, each paying a ...

Feb 16, 2009 · A bus company has 4000 passengers daily, each paying a a fare of For each each $0.15 increase, the company estimates that it will lose 40 passengers. If the company needs to take in $10 450 per day to stay in business, what fare asked by Mark on May 19, 2018…

## A bus company has 4000 passengers daily each paying a fare ...

exam Numerical Ability Question Solution - A bus company has 4000 passengers daily, each paying a fare of $2. For each $0.15 increase, the company estimates that it will lose 40 pasengers per pay. If the company needs to ta…

## A bus company has 4000 passengers daily, each paying a ...

Oct 27, 2011 · A bus company has 4000 passengers daily, each paying a fare of $2. For each $0.15 increase, the company estimates that it will lose 40 passengers per day. If the company needs to take in $10 450 per day to stay in business, what fare should be charged? Show me your steps please.…

## A bus company has 4000 passengers daily, each paying a ...

Feb 17, 2007 · Ok, so 4000 x 2 is 8000. So the company still needs 2450 to meet the criteria. So need to charge more. If they did the 0.15 increase 40 people will bugger off so 3960 x 2.15 is 8514 so they still need more. Again passengers leave with the charge increased so 3920 x 2.30 is 9016.…

## A bus company has 4000 passengers daily each paying a fare ...

A bus company has 4000 passengers daily, each paying a fare of $2. For each $0.15 increase, the company estimates that it will lose 40 passengers per day. If the company needs to take in $10 450 per day to stay in business, what fare should be charged? 15. Describe three possible ways that you could determine the zeros of the quadratic function 16.…