4000 passengers @ $2; 40 fewer for each $0.15 increase; find

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vedmachka

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This is due tomorrow and I am at a loss. I know how to solve these problems, but this one just got me stuck for some reason. Any help would be great.

"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. What fare should the company charge to maximize their revenue?"

Now I know the answer already and solved this problems using derivatives and the following equation:

(4000-40x)(2+0.15x)=0

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 can substitute a value into the dependent variable.

Any help would be greatly appreciated... I don't know how to solve it other than what I already did...
 
Re: Easy problem, but got me completely 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 15-cent fare increases, then the daily revenue is a function of x.

R(x) = (4000 - 40x)*(2 + 0.15x)

Do you recognize this as a quadratic? 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.

x-coordinate of vertex = -b/(2a)

Once you know the x-value that generates the maximum revenue, you can reason as to how to round it to figure out how many 15-cent increases to add to $2 to answer the question.

(I'm not sure what you mean by "substitute a value into the dependent variable", since R(x) is the dependent variable. If substituting anything, it seems to me that you want to substitute the value of x at the vertex into the expression 2+0.15x to determine the fare.)

Cheers,

~ Mark :)

MY EDIT: added comment pertaining to the fact that the value of x at the vertex must be properly rounded to get a fare increase that is a multiple of 15 cents.
 
Re: Easy problem, but got me completely lost....

Don't know what to tell you; your method is certainly valid, and results (rounded) in 2280 @ 8.45 = 19,266 : correct?
 
Re: Easy problem, but got me completely lost....

Denis said:
Don't know what to tell you; your method is certainly valid, and results (rounded) in 2280 @ 8.45 = 19,266 : correct?
Click to expand...

Hmmm, I arrived at $8.50 for the fare, Denis.

I double-checked my arithmetic; Is my posted strategy okay?

~ Mark :?
 
Re: Easy problem, but got me completely lost....

I restricted the increase to multiples of .15: 2.15, 2.30, ... 8.30, 8.45, 8.60, ...
 
Re: Easy problem, but got me completely lost....

vedmachka said:
Ok,
This is due tomorrow and I am at a loss. I know how to solve these problems, but this one just got me stuck for some reason. Any help would be great.

"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. What fare should the company charge to maximize their revenue?"

Now I know the answer already and solved this problems using derivatives and the following equation:

(4000-40x)(2+0.15x) = 0.........????

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.

\(\displaystyle R(x) \, = \, (4000 - 40x)\cdot (2 \, + \, 0.15x)\)

for maximum of R(x), we have,

\(\displaystyle \frac{dR(x)}{dx} \, = \, 0\)

\(\displaystyle 600 - 80 - 12x = 0\) .................... for maximum revenue

\(\displaystyle \text x \, = \, \frac{520}{12} \, = 43^{+} \, = \, 43\)

New fare = 2 + 43 * .15 = 8.45

In reality though, if you increase that price in one shot - there will be riot in your hand and you'll lose all the ridership.

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 can substitute a value into the dependent variable.

Any help would be greatly appreciated... I don't know how to solve it other than what I already did...
Click to expand...
 
Re: Easy problem, but got me completely lost....

Thank you guys for your replies, however that is the solution that I got as well. The answer is $8.45 for the fare. I did the question using derivatives. The teacher said not to use derivatives. Still lost....
 
Re: Easy problem, but got me completely lost....

Denis said:
I restricted the increase to multiples of .15: 2.15, 2.30, ... 8.30, 8.45, 8.60, ...
Click to expand...

Hi Denis:

DOH! :oops:

Coincidentally, the last time I made this type of goof, it was also with buses. I once reported that the army needed something like 13.2 buses to move some troops ...

Darn it! Oh well. (I recently posted that mathematics proceeds by making mistakes. The way this week's been going for me, I must really be proceeding ... :lol: )

~ Mark
 
Re: Easy problem, but got me completely lost....

vedmachka said:
... Still lost....
Click to expand...

Hello Vedmachka:

Since you seem to understand enough calculus to know what you're doing, I'm curious to know why you're still lost.

Do you realize that the revenue function is defined by a quadratic expression?

Do you remember that the graph of a quadratic is a parabola?

All parabolas have a vertex, and, if the parabola is the graph of a function, then the y-coordinate of this vertex is the function's maximum (or minimum) value because the vertex is always the highest (or lowest) point on the graph.

If you would like me to upload a graph of the revenue function in your exercise, then I will be glad to do it.

The x-coordinate of the vertex is the value of x that maximizes the revenue. (The vertex is THE point on the graph where a tangent line has slope zero, so its x-coordinate is the same value of x that you arrived at using calculus.)

Therefore, you can get the same result without using calculus by simply discovering the x-coordinate of the vertex, followed by a little reasoning involving the fact that the fare increase must be a multiple of 15-cents. (This reasoning, as explained by Denis, is the part that I forgot to do.)

I explained how to discover the x-coordinate of the vertex in my first post, but I'll repeat it.

You have a choice.

You can use the standard formula for the x-coordinate of the vertex.

Given the form y = ax^2 + bx + c, the x-coordinate of the vertex is -b/(2a).

OR, you can use the quadratic formula to get the two roots (these values correspond to the two x-intercepts for this particular graph). Because the parabola is symmetrical, the x-coordinate of the vertex lies halfway between the x-intercepts.

If you're still lost after thinking about parabolas, then please let us know. It's not difficult to understand, and I'll be happy to provide a detailed picture of the graph in your particular exercise.

Cheers,

~ Mark :)
 
Re: Easy problem, but got me completely lost....

mmm4444bot said:
... OR, you can use the quadratic formula to get the two roots ...
Click to expand...

SCRATCH THAT!!

No need for the quadratic formula to find the roots, since you've already got a factored form.

Set each factor equal to zero, and solve for x.

4000 - 40x = 0

2 + 0.15x = 0

If the two roots are R1 and R2, then the x-coordinate of the vertex is (R1 + R2)/2.

So, this method is the better choice because you would first need to multiply the factors to calculate the values of a and b if you wanted to use -b/(2a).

Cheers,

~ Mark :)

(Rhetorical questions: Why do I always gravitate to the "hard" way? Why do I always choose the slowest line at the grocery store?)
 
Re: Easy problem, but got me completely lost....

My teacher writes the following: "Solutions that are not grade 11 appropriate are, for example, making a table of values, using derivatives, using the (–b/2a) ‘formula’. Even though those solutions may be easier, and may also produce the correct answer, they are not in the grade 11 curriculum." I am supposed to do this: "When dealing with a max/min question, a grade 11 solution would involve modeling the problem with a quadratic function, then determining the vertex by completing the square, factoring to find the zeros and then averaging them, or finding two symmetrical points (one of which being the y-intercept) and then averaging them." However, maybe knowing calculus too well, it is bad for me. I can't seem to solve this any other way.

However, I have completed the square and got x=100, -40/3
lost from here on....
 
Re: Easy problem, but got me completely lost....

vedmachka said:
... factoring to find the zeros and then averaging them ...

... I have completed the square and got x=100, -40/3
lost from here on....
Click to expand...

You're oh so close!

R1 = 100

R2 = -40/3

"Averaging them" means (R1 + R2)/2

You should end up with the same value that you discovered using calculus.

~ Mark :)
 
Re: Easy problem, but got me completely lost....

THANK YOU SOOOOO MUCH MARK!!!!!!! :D YOU ARE THE BEST!!!!!!!! I FIGURED IT OUT A SECOND BEFORE YOUR POST.... :!:
 
Re: Easy problem, but got me completely lost....

vedmachka said:
... YOU ARE THE BEST!!!!!!!! This statement is clearly false, but I'm glad you're happy. :wink:
Click to expand...
 
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