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Math Help - Need help with parabolas and such.. afew questions

  1. #1
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    Need help with parabolas and such.. afew questions

    I've been practically falling asleep in class for the past week and now I need to study my ass off to make up for it. Here are 3 questions that I'm having trouble with.

    1) Given y = -4(x-6)^2+5
    a) Find the maximum or minimum vaue of y. Which is it?
    b) For what value of x does the max or min occur?

    2) A rectangle has dimensions 3x and 5-2x
    a) What is the maximum area of the rectangle?
    b) What value of x gives the maximum area?

    3) A rectangular lawn measures 7m by 5m (I'm in canada, we do meters!) A uniform border of flowers is to be planted along two adjacent sides of the lawn. If the flowers that have been purchased will cover an area of 6.25m^2, how wide is the border?

    Thanks alot!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Sculthorp View Post
    I've been practically falling asleep in class for the past week and now I need to study my ass off to make up for it. Here are 3 questions that I'm having trouble with.

    1) Given y = -4(x-6)^2+5
    a) Find the maximum or minimum vaue of y. Which is it?
    the coefficient of x^2 is negative, so it is a downward opening parabola, it has a max.

    this occurs at (6,5) -- do you see why?
    b) For what value of x does the max or min occur?
    the answer is my last response

    since (x - 6)^2 is nonnegative, we have the max when we subtract as little as possible, that is, when (x - 6)^2 = 0
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Sculthorp View Post
    2) A rectangle has dimensions 3x and 5-2x
    i suppose the dimensions are (height) and (base)?

    a) What is the maximum area of the rectangle?
    if so, use the fact that \mbox { Area } = \frac 12 (\mbox { base })( \mbox { height })

    find the max value of the resulting quadratic. you can use my last post to guide you once you've completed the square

    b) What value of x gives the maximum area?
    again, see my last post
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Sculthorp View Post
    3) A rectangular lawn measures 7m by 5m (I'm in canada, we do meters!) A uniform border of flowers is to be planted along two adjacent sides of the lawn. If the flowers that have been purchased will cover an area of 6.25m^2, how wide is the border?
    see the diagram below.

    Let x be the width of the border

    then the dimensions of the lawn besides the border are: length = 7 - x and width = 5 - x

    Now, \mbox { Area of the whole lawn } = 7 \times 5 = 35 \mbox { m}^3

    we are told, \mbox { Area of flower garden } = 6.25 \mbox { m}^3

    So, \mbox{ Area of lawn besides the flower garden } = 35 - 6.25 = 28.75 \mbox { m}^3

    Thus, \mbox {Area of lawn besides the flower garden } = (7 - x)(5 - x) = 28.75

    solve that quadratic for x and you're done
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    Okay all of the answers were great except for the last one.. even though you even made a diagram! (5+x)(7+x) is the area not (5-x)(7-x) if you know what I'm getting at. The answer ends up being 0.5m somehow (says the back of the book). The original box is 7x5 and the flowers are added. So the last equation is (7+x)(5+x)=41.25, but how do I solve that for x?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Sculthorp View Post
    Okay all of the answers were great except for the last one.. even though you even made a diagram! (5+x)(7+x) is the area not (5-x)(7-x) if you know what I'm getting at. The answer ends up being 0.5m somehow (says the back of the book). The original box is 7x5 and the flowers are added. So the last equation is (7+x)(5+x)=41.25, but how do I solve that for x?
    ah, ok. it seems i misinterpreted the question. i thought the flower garden was a part of the lawn not an addition to it (since that makes no sense, if you're planting flowers outside of the width of your lawn, chances are you are doing it on someone else property). but anyway, it seems you corrected the mistake. good job altering my solution! it's an indication that you actually understand what's going on

    to solve the new equation, just expand the left hand side and simplify to get a quadratic.

    (7 + x)(5 + x) = 41.25

    \Rightarrow x^2 + 12x + 35 = 41.25

    \Rightarrow x^2 + 12x - 6.25 = 0

    now solve that (use completing the square or the quadratic formula)
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  7. #7
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    x^2+12x+36=6.25+36
    (x+6)^2=42.25
    x+6=6.5
    x=0.5



    You're awesome Jhevon! My first day registered here and I'm already planning on donating when I get money in paypal
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Sculthorp View Post
    x^2+12x+36=6.25+36
    (x+6)^2=42.25
    x+6=6.5
    x=0.5



    You're awesome Jhevon! My first day registered here and I'm already planning on donating when I get money in paypal
    oh, no. i did not do much for you. you just needed a nudge in the right direction. you're algebra is good, which was more than half the work here. good luck with your class

    a caution though: you should have gotten two answers here, one negative and one positive, then you would reject the negative one, since \sqrt {42.25} = \pm 6.5
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