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Math Help - Parabolas in word problems

  1. #1
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    Parabolas in word problems

    Hi guys!

    We gad a disscusiion on parabolas last week and I wasn;t able to understand it very well . Over the weekend,, our teacher gave us 35problems to solve, but I was stuck on these four numbers... How do you solve them? I don't think we have studied them yet in our class. Thanks!!

    1) Find 2 numbers whose diff. is 20 and whose product is a minimum

    2) What are the dimensios if the largest rectangle with a fixed perimetre "P"?

    3) A student club on a collegre campus charges an annual membership dues of $40, less 20cents for each member over 60 (no. of ppl). How many members will give the clubthe most revenue from annual duies? What is the max. revenue??

    4) A travel agency offers an organization an all inclulive tour for $8000 per person if not more than 100 persons join the tour. The cost per person will be reduced by $50 for each person in excess of 100. Find the largest possible gross sales revenue for the agency, and how many persons should join the tour? What's the cost per person so that the max revenue is achieved??
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by AndrewK View Post
    Hi guys!

    We gad a disscusiion on parabolas last week and I wasn;t able to understand it very well . Over the weekend,, our teacher gave us 35problems to solve, but I was stuck on these four numbers... How do you solve them? I don't think we have studied them yet in our class. Thanks!!

    1) Find 2 numbers whose diff. is 20 and whose product is a minimum
    I assume you're in precalculus, correct? i'll try to keep the solutions as simple as possible.

    let the two numbers be x and y, we have:

    x - y = 20
    P = xy ..........where P is the product

    since x - y = 20
    => x = 20 + y
    substitute 20 + y for x in P, we get:

    P = (20 + y)y
    => P = 20y + y^2
    this is an upward opening parabola, the minimum value is at the vertex. the vertex occurs when:
    y = -b/2a , where b is the coefficient of y and a is the coefficient of y^2

    so for vertex, y = -20/2 = -10

    but x - y = 20
    => x + 10 = 20
    => x = 10

    so the two numbers are 10 and -10


    2) What are the dimensios if the largest rectangle with a fixed perimetre "P"?
    let the length be x
    let the width be y

    the largest rectangle will have the largest area


    then P = 2x + 2y where P is the perimeter

    now A = xy

    since P = 2x + 2y
    => y = (P - 2x)/2
    substitute (P - 2x)/2 for y in A, we get:

    A = x(P - 2x)/2
    => A = (1/2)(Px - 2x^2) = (P/2)x - x^2
    this is a downward opening parabola, the max value occurs at the vertex. for the vertex, x = -b/2a
    => x = -(P/2)/-2 = P/4

    but y = (P - 2x)/2
    => y = (P - 2(P/4))/2
    => y = (P - P/2)/2
    => y = (P/2)/2
    => y = P/4

    so the largest rectangle with a fixed perimeter P, is a square with dimentions length = width = P/4
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  3. #3
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    Jhevon,

    Yup, I am in precalculus.

    I tried doing no. 3 again, and came up with a negative ans (-145.9867 something ???). I just can't figure out what I did wrong in my solution
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  4. #4
    Bar0n janvdl's Avatar
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    Hmm, tricky questions i must say.

    For number 3:

    R = 40x + 39,8y

    Where R is the revenue ; x the amount of people paying $40 ; and y the amount of people over 60.

    I'm not sure how to take it further...
    Last edited by janvdl; May 6th 2007 at 04:27 AM.
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  5. #5
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    Quote Originally Posted by AndrewK View Post
    Hi guys!

    ...

    3) A student club on a collegre campus charges an annual membership dues of $40, less 20cents for each member over 60 (no. of ppl). How many members will give the clubthe most revenue from annual duies? What is the max. revenue??
    ...
    Hello, AndrewK,

    if I understand this problem correctly, then each of the first 60 students pays $40. Student #61 pays $39.80, student #62 pays $39.60, etc.

    If x is the number of students the revenue of the club will be:

    r = 60 * $40 + (x - 60)(40 - 0.2*x) with x >= 60

    r = -0.2x + 52x

    This is the equation of a parabola opening downward. The maximum value is at it's vertex. (To calculate the vertex compare the posts of Jhevon):

    x = 130 will give r = $3380
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  6. #6
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    Quote Originally Posted by AndrewK View Post
    Hi guys!

    ...

    4) A travel agency offers an organization an all inclulive tour for $8000 per person if not more than 100 persons join the tour. The cost per person will be reduced by $50 for each person in excess of 100. Find the largest possible gross sales revenue for the agency, and how many persons should join the tour? What's the cost per person so that the max revenue is achieved??
    Hello, AndrewK,

    let x be the number persons and x >= 100

    price = $8000 - $50(x-100)

    then the revenue is:

    r = x(8000 - 50(x-100))

    r = -50x + 13000x

    This is the equation of a parabola opening downward. The maxiumum value is at it's vertex. thus:

    x = 130 and r_max = $845,000
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  7. #7
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    Hello, Andrew!

    I have a slightly different set-up for #3 and #4.
    . . I think it's simpler, but that's just my opinion.


    3) A student club on a collegre campus charges an annual membership dues of $40,
    less 20 cents for each member over 60 members.
    How many members will give the club the most revenue from annual duies?
    What is the max. revenue?

    Let x = number of members in excess of 60.
    . . Then there will be 60 + x members.

    Each will pay a reduced fee: .40 - 0.20x dollars.

    The revenue is: .R .= .(60 + x)(40 - 0.20x) .= .2400 + 28x - 0.2x

    This is a down-opening parabola; its maximum is at its vertex.
    The vertex of parabola, y .= .ax + bx + c .is at: .x .= .-b/2a

    We have: .a = -0.2, b = 28, c = 2400

    The vertex is at: .x .= .-28 2(-0.2) .= .70


    Therefore, maximum revenue is achieved with: .60 + 70 .= .130 members

    And the revenue will be: .R .= .2400 + 28(70) - 0.2(70) .= .$3,380



    4) A travel agency offers an organization an all inclulive tour for $8000 per person
    if not more than 100 persons join the tour.
    The cost per person will be reduced by $50 for each person in excess of 100.
    Find the largest possible gross sales revenue for the agency,
    and how many persons should join the tour?
    What's the cost per person so that the max revenue is achieved?

    Let x = number of people in excess of 100.
    . . There are 100 + x people.

    Each gets a reduced fare: .8000 - 50x dollars

    The revenue is: .R .= .(100 + x)(8000 - 50x) .= .800,000 + 3000x - 50x

    This is a down-opening parabola with its maximum at its vertex.
    We have: .a = -50, b = 3000, c = 800,000
    . . Hence, the vertex is at: .x .= .-3000/2(-50) .= .30


    Therefore, for maximum revenue:
    . . 100 + 30 .= .130 people should join the tour,
    . . who will pay 8000 - 50(30) .= .$6500 each,
    . . for a maximum revenue of: .130 $6500 .= .$845,000

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  8. #8
    Bar0n janvdl's Avatar
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    Thank you Earboth and Soroban.
    I too understand these problems better now.
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