Grade 10 Quadratic Math Help?

Question

How would I solve this. It'd be helpful if you showed your work.

A bus company has 4000 passengers daily,each paying a fare of 2$. For each 0.15$ increase the company estimates it will lose 40 passengers. If the company needs to take in 10 450 per day to stay in business what fare should be charged.

Answer 1

Hi,

Let x = number of 15 cent increases made to the price

Then 2 + .15x is the new price and 4000 - 40x is the number of passengers.

The quadratic is Y = (2 + .15x)(4000 - 40x). It has a maximum at x = 43.333 and y = $19,266.67 as the maximum income if the price is increased to $8.50 per passenger and the number of passengers decreases to 2266.66 passengers. Since you can't have part of a passenger, we will revise this to a price of $8.45 for 4000 - 40(43) or 2,280 passengers. This gives a maximum income of $19,266.

If they want to make the minimum of $10,450 at the lowest per person cost, they should increase the price by five .15 to $2.75. This would give 3800 people rides at the lowest possible cost to them.

I hope that helps!! :-)

Answer 2

$2.75