Algetbra grade 10 math

Question

A bus company usually transports 12000 people per day at a ticket price of 1$. The company wants to raise the ticket price. For every 0.10$ increase in the ticket price the number of riders is expected to decrease by 400. Calculate the ticket price that will max revenue

Answer 1

So your income function can be represented by

 

R(x) = (12000 - 400x)(1+.1x) where x is the number of increases in price. check out that your function is correct first and make sense. If there is 1 increase, the riders go down by 400 and the price went up by 10 cents. 2 increases would lose 800 passengers and the price would be $1.20, an increase of 20 cents....so this function works.

 

When you multiply it out, you get a parabola opening downward so that it has a maximum value which will be on the axis of symmetry. plug in that value and you get the turning point where x is the number of increases and y is the maximum revenue (income).

 

R(x) = 12000 + 800x - 40x²

The axis of symmetry is x= -b/2a so you get x=-800/-80 = 10.

plug in 10 to get R(10) = 12000 + 8000 - 4000 = 16000.

 

So (10,16000) is your turning point meaning 10 increases will lead to a maximum revenue or $16000. You can also check this on a graphing calculator.

 

I'm also going to check by finding R(9) and R(11). These should come out the same since they are both one away from the axis of symmetry and they're on opposite sides of it. Thgey shoud both also be less than 16000.

R(9) = 12000 + 800*9 - 40*81 = 15960.

R(11) = 12000 + 800*11 - 40(121) = 15960.

 

So I'm convinced that my answer is correct.

 

Hope this helped.