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 Another simple quadratic question!!!
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Nathan4102




PostPosted: Sun May 12, 2013 5:26 pm   Post subject: Another simple quadratic question!!!

Alright, so I'm working on more of this quadratics work, and I've come across another problem. Its not that I can't solve this one, its that I think I'm doing it the wrong way. It took me about 3 minutes to find the equations for the let statements, and a decent amount of time to solve the equation because of all the fractions. Obviously, I need all the time I can get on tests and exams, so is there a cleaner, better, and faster method to solve this?

Q: A bus company carries about 20 000 riders per day for a fare of $0.90. A survey indicates that if the fare is decreased, the number of riders will increase by 2000 for every decrease of $0.05. Find the ticket price that will result in the greatest revenue.

A:
CODE:

Note: 56 000 - 40 000x = 20 000 + 40 000(.9 - x)

let x rep the bus fare.
let 56 000 - 40 000x rep the amount of riders riding the bus.
let y rep the total daily revenue.
y = (x) (56 000 - 40 000x)
y = 56 000x - 40 000x^2
y = -40 000 (x^2 - 7/5x)
y = - 40 000 (x^2 - 7/5x + 49/100) - 49/4 000 000
y = -40 000 (x - 7/10)^2 - 49/4 000 000

7/10 = 0.7

Therefore the ticket price should be $0.70 in order to achieve maximum daily revenue.
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whoareyou




PostPosted: Sun May 12, 2013 6:06 pm   Post subject: RE:Another simple quadratic question!!!

When you're doing quadratic optimization, most of the time you'll need to find the vertex of the function. The x-value of the function is -b/2a when the equation is in the form ax^2 + bx + c. So at the second line after your let statements, -56000/-80000 = 0.7.
TZak




PostPosted: Sun May 12, 2013 6:06 pm   Post subject: Re: Another simple quadratic question!!!

Find the derivative.
whoareyou




PostPosted: Sun May 12, 2013 6:08 pm   Post subject: RE:Another simple quadratic question!!!

Sorry, "The x-value of the function is" should read "The x-value of the vertex is."
Nathan4102




PostPosted: Sun May 12, 2013 6:11 pm   Post subject: Re: RE:Another simple quadratic question!!!

whoareyou @ Sun May 12, 2013 7:06 pm wrote:
When you're doing quadratic optimization, most of the time you'll need to find the vertex of the function. The x-value of the function is -b/2a when the equation is in the form ax^2 + bx + c. So at the second line after your let statements, -56000/-80000 = 0.7.

"-b/2a"? Not too sure what you mean by this. This is only grade 10, btw, in case thats 11+.

TZak wrote:

Find the derivative.


We haven't been taught derivatives yet. I could learn them on my own, but my teacher wouldn't mark it.
Nathan4102




PostPosted: Sun May 12, 2013 6:12 pm   Post subject: Re: RE:Another simple quadratic question!!!

whoareyou @ Sun May 12, 2013 7:08 pm wrote:
Sorry, "The x-value of the function is" should read "The x-value of the vertex is."


Ok, now the -b/2a makes a bit more sense, but could you explain how you got it?
Zren




PostPosted: Sun May 12, 2013 6:14 pm   Post subject: RE:Another simple quadratic question!!!

Where did 56 000 come from?

I interpreted it as:

Revenue = Riders * Cost
Revenue = 20000 * $0.90

let y represent revenue
let x represent ______ <= Not quite sure what to call it really.
y = (20000 + 2000x)($0.90 - $0.05x)

It's in factored form. So you just have to convert it to vertex form to get x at the maximum/vertex of the parabola.

I don't quite remember how to write vertex form so...

http://www.wolframalpha.com/input/?i=%2820000+%2B+2000x%29%280.90-0.05x%29

Which tells me the maximum is at x=4. Which means Cost=$0.90*$0.05(4) = $0.70. So your answer is correct... somehow.
whoareyou




PostPosted: Sun May 12, 2013 6:20 pm   Post subject: RE:Another simple quadratic question!!!

A quadratic equation in standard form has the general formula Ax^2 + Bx + C.

I'm stating that the x-value of vertex of the quadratic function in standard form is given by -B/2A (ie. the coefficients in the general form).

For example, y = 56 000x - 40 000x^2 is in standard form. So to get the vertex, you identify B = 56000 and A = -40000, so -B/2A = -56000/2(-40000) = 0.7.

This is actually part of the quadratic formula, so you should have learned it. However there is a simple way to derive that formula for the vertex. If you have the quadratic in standard form, you take the derivative and set it equal to zero (ie. the rate of change at the vertex is zero).

So d/dx (Ax^2 + Bx + C) = 2AX + B.
0 = 2Ax + B
-B = 2Ax
x = -B / 2A
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Nathan4102




PostPosted: Sun May 12, 2013 6:23 pm   Post subject: Re: RE:Another simple quadratic question!!!

Zren @ Sun May 12, 2013 7:14 pm wrote:
Where did 56 000 come from?

I interpreted it as:

Revenue = Riders * Cost
Revenue = 20000 * $0.90

let y represent revenue
let x represent ______ <= Not quite sure what to call it really.
y = (20000 + 2000x)($0.90 - $0.05x)

It's in factored form. So you just have to convert it to vertex form to get x at the maximum/vertex of the parabola.

I don't quite remember how to write vertex form so...

http://www.wolframalpha.com/input/?i=%2820000+%2B+2000x%29%280.90-0.05x%29

Which tells me the maximum is at x=4. Which means Cost=$0.90*$0.05(4) = $0.70. So your answer is correct... somehow.


In our class, x almost always represents whats being looked for, in this case, the price. and since y is the multiplication of x and the other variable, the other variable has to be number of people. That 56000 thing was just (20 000 + 40 000(.9 - x)) simplified, which is my equation for number of people riding the bus. I'll have to check with my teacher weather or not I'm allowed to use your method. He's usually pretty set on us using the methods taught though. Thanks for the explanation
Nathan4102




PostPosted: Sun May 12, 2013 6:27 pm   Post subject: Re: RE:Another simple quadratic question!!!

whoareyou @ Sun May 12, 2013 7:20 pm wrote:
A quadratic equation in standard form has the general formula Ax^2 + Bx + C.

I'm stating that the x-value of vertex of the quadratic function in standard form is given by -B/2A (ie. the coefficients in the general form).

For example, y = 56 000x - 40 000x^2 is in standard form. So to get the vertex, you identify B = 56000 and A = -40000, so -B/2A = -56000/2(-40000) = 0.7.

This is actually part of the quadratic formula, so you should have learned it. However there is a simple way to derive that formula for the vertex. If you have the quadratic in standard form, you take the derivative and set it equal to zero (ie. the rate of change at the vertex is zero).

So d/dx (Ax^2 + Bx + C) = 2AX + B.
0 = 2Ax + B
-B = 2Ax
x = -B / 2A


My god... That would make everything so much simpler! I'm not sure why we haven't been taught this... I'm going to ask my teacher if I can use this method. Thanks for this method, if I'm allowed to use this, its going to save me sooo much time! The method we use now takes forever!
Dreadnought




PostPosted: Sun May 12, 2013 8:03 pm   Post subject: Re: Another simple quadratic question!!!

I can't see why your teacher would be against this, there is no math needed that is is beyond the scope of grade 10.

When you're trying to minimize a parabola opening upwards, or maximize one opening downwards, the optimal point is clearly the trough or the peak respectively. Since parabolas are symmetric, the peak or trough is halfway between the two x-intercepts. From the quadratic formula it should be obvious that this midpoint is -b/2a (like whoareyou said).

Symmetry is very nice and can often save you a little time.

Note: Don't worry about cases where the function doesn't cross the x-axis since the complex roots (have you learned imaginary numbers yet?) will also cancel each other.
Panphobia




PostPosted: Sun May 12, 2013 8:12 pm   Post subject: RE:Another simple quadratic question!!!

Everything I think from grade 7 or 8 has been preparing you for calculus and all of its elegance. So right now I am guessing you are in grade 10, you are doing quadratic optimization the ugly and barbaric way. Just wait until grade 12 calculus, everything will be so simple and fun. I am afraid you are going to be stuck with this ugly way up until calculus.
Nathan4102




PostPosted: Sun May 12, 2013 9:55 pm   Post subject: RE:Another simple quadratic question!!!

Imaginary numbers are next year dread, I guess ill just have to do it this way for now.

Thanks pan, I think youre the first person ive seen say something positive about 12 calculus! All ive heard is "its hell!" "Dont take it" "Calc made me hate math" blah blah blah :p
Dreadnought




PostPosted: Sun May 12, 2013 10:36 pm   Post subject: Re: Another simple quadratic question!!!

Sorry, that question was just out of curiosity, it doesn't really help with this problem. Don't worry about imaginary numbers.

The important thing is that a parabola is symmetric about a vertical line through its vertex (peak/trough) and that from this we can see that the midpoint of the x-intercepts of the parabola is the x-coordinate of the vertex. Then we can use the x-intercepts given by the quadratic formula to show the midpoint is at x = -b/2a, regardless of whether the function has real x-intercepts or not. (So even though you no longer have x-intercepts on a graph of your function, the math will still work out)

Also, calculus is very nice. If you have any interest in science or math you really should take it, and if you have any interest in math you'll probably enjoy it too!
Nathan4102




PostPosted: Mon May 13, 2013 9:02 am   Post subject: RE:Another simple quadratic question!!!

Yeah, x = -b/2a makes sense to me. Hopefully we get to use it soon. I hate this long process just to find something visible in the second line. I have to take calculus for University compsci, so hopefully it will be good!
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