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Math Help - maximum value of area

  1. #1
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    maximum value of area

    A rectangle with one side on the x-axis and one side on the line has its upper left vertex on the graph of y=x^2 For what value of x does the area of the rectangle attain its maximum value?

    I know the answer is 4/3.

    Do you solve this by rectangle approximation?
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  2. #2
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    Re: maximum value of area

    Quote Originally Posted by Jonroberts74 View Post
    A rectangle with one side on the x-axis and one side on the line has its upper left vertex on the graph of y=x^2 For what value of x does the area of the rectangle attain its maximum value?

    I know the answer is 4/3.

    Do you solve this by rectangle approximation?
    No need for any approximations here. If I understand your setup you have a rectangle with vertices at

    (0,0) (0,x2), (-x, x2), (-x,0)

    this has area of -x3

    As x -> -infinity this area grows w/o bound so there is no maximum.

    Perhaps you left something out of the problem statement?
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  3. #3
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    Re: maximum value of area

    ah it should say with one side at the x-axis and one side at the line x = 2 and the upper left vertex on the graph y=x^2.

    sorry
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    Re: maximum value of area

    this is still infinite area as x -> -infinity
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  5. #5
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    Re: maximum value of area

    it doesn't give infinity as an answer, it is multiple choice with the answers

    2, 4/3 , 1, 3/4 and 2/3

    and according to the answer key the correct one is x=4/3
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    Re: maximum value of area

    Quote Originally Posted by Jonroberts74 View Post
    it doesn't give infinity as an answer, it is multiple choice with the answers

    2, 4/3 , 1, 3/4 and 2/3

    and according to the answer key the correct one is x=4/3
    Well it's pretty clear that if you have a rectangle with it's right side at x=2, and it's bottom at y=0, and let the left side go off to -infinity that you're going to have an infinite area rectangle. You can see that can't you?

    Ok, I see it. x is restricted to x >= 0.

    So what have you got. Your vertices are at (x,0), (x,x^2), (2,x^2), (2,0)

    length is 2-x, width is x^2, area is (2-x)x^2

    Use your calculus to find the maximum of this function. Take it's derivative and set it equal to 0 and solve for x.

    One of these values is a maximum and one will be a minimum. Use the 2nd derivative to figure out which is which (though it should be pretty obvious from their values).
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  7. #7
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    Re: maximum value of area

    Hello, Jonroberts74!

    A rectangle with one side on the x-axis and one side on the line x=2
    has its upper left vertex on the graph of y=x^2.
    For what value of x does the area of the rectangle attain its maximum value?

    Did you make a sketch?

    Code:
                  |                .
        *         |         *      . 
                  |                . 
         *        |        *-------*
          *       |       *|:::::::|y
             *    |    *   |:::::::|
        ----------*--------*-------*---
                  |    x   :  2-x  2 
                  |
    The area of the rectangle is: . A \;=\;y(2-x) \;=\;x^2(2-x)

    You must maximize the function: . A \;=\;2x^2 - x^3

    Got it?

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  8. #8
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    Re: maximum value of area

    did you do that in a text editor? good lord.
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  9. #9
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    Re: maximum value of area

    got it, thank you!
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  10. #10
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    Re: maximum value of area

    Hi,
    I'm with Soroban. You should sketch a picture. I enjoy drawing pictures with "modern" software. Two advantages are evident. One, the graphs can be accurate in the sense that the true shape of the graph is drawn. Before the advent of modern software, I didn't know what the graph of y = x2 really looks like. Secondly, in the static drawing attached, it comes from a program that allows me to have a slider for x. I can then dynamically change x and see the various rectangles; also this allows me to check and see if I have found the maximum area.

    maximum value of area-mhfcalc23.png
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