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Math Help - Solving Dimensions given Area

  1. #1
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    Solving Dimensions given Area

    A page contains 32 square inches of print. The margins at the top and the bottom of the page are 2 inches. The margins on each side are only 1 inch. Find the dimensions of the page so that the least paper is used.

    I was taught to write two equations, one for perimeter and one for area.
    I know that the equation for area is A=(x-2)(y-4)=32. Since I want the least paper used, I need to find the minimum of perimeter. But I don't know what I should put for the equation.


    A rectangle is bounded by the x-axis and the semicircle y=\sqrt{16-x^2}. What length and width should the rectangle have so that its area is at a maximum?

    I'm not sure how to show you the image, but its a graph with a semicircle on top of the x-axis in the middle of the y-axis with radius reaching out to (-4,0) and (4,0) from the origin. There's a rectangle within this semicircle.
    So the length is split by the y-axis forming an x on each side. The width, y, stays the same. The area equation I wrote was A=2xy=max which is correct. Since I'm given y, I replace it in the equation giving me: A=2x(\sqrt{16-x^2}), and take the derivative to get A'=\frac{-2x^2}{\sqrt{16-x^2}}+2\sqrt{16-x^2}. So I set it equal to zero, and get the values x=-2\sqrt{2} and x=2\sqrt{2}. But my teacher tells me there are more x values than that.

    Thank you for your time.
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  2. #2
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    For the first problem, it might help to draw a diagram. You are told that on a printed page the dimensions of PRINT (the words) is 32 square inches:

    32=x\cdot y

    We are also told that this page that printed text lies on, has 2 inch margins on the top, and 1 inch margins on the side. Thus the dimensions of the page are:

    l=x+2; w=y+4

    A_{entire page}=(x+2)(y+4)

    Now in order to make a clean equation for the area of the ENTIRE page (not the printed area), we need to create an equation in a single variable. Going back to the first equation:

    32=x\cdot y \Longrightarrow \frac{32}{x}=y

    Now we can plug into our equation for the area of the entire page:

    A_{entire page}=(x+2)(\frac{32}{x}+4) \Longrightarrow \\* A_{entire page}=(x+2)(\frac{32+4x}{x}) \Longrightarrow \frac{4x^2+40x+64}{x}

    Now we need to take the derivative of our function for the area of the entire page:

    A'_{entire page}=\frac{(8x+40)x-(4x^2+40x+64)(1)}{x^2} \Longrightarrow A'_{entire page}=\frac{8x^2+40x-4x^2-40x-64}{x^2} \Longrightarrow A'_{entire page}=\frac{4x^2-64}{x^2}

    From here we set the equation equal to zero, to find the minimum:

    0=\frac{4x^2-64}{x^2} \Longrightarrow 64=4x^2 \Longrightarrow \pm 4=x

    One of these solutions, is obviously not possible. From here it is simply a matter of plugging the value of X into our original equation to find the length equal to 6. Plugging X into the equation we used to eliminate Y allows us to find that dimensions of the printed work to be 8, and thus the dimensions of the entire pages width to be equal to 12.
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  3. #3
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    Hello, DarkestEvil!

    A page contains 32 inē of print.
    The margins at the top and the bottom of the page are 2 inches.
    The margins on each side are only 1 inch.
    Find the dimensions of the page so that the least paper is used.

    I was taught to write two equations, one for perimeter and one for area.
    No, that is for a specific (and different) problem.
    Code:
          : 1 : - x - : 1 : -
        _ *---------------*  -
        : |               |  :
        2 |               |  :
        : |               |  :
        - |   *-------*   |  :
        : |   |///////|   |  :
        : |   |///////|   |  :
        y |   |///////|y  | y+4
        : |   |///////|   |  :
        : |   |///////|   |  :
        - |   *-------*   |  :
        : |       x       |  :
        2 |               |  :
        : |               |  :
        - *---------------*  -
          : - -  x+2  - - :

    Let: x = width of printed area.
    Let: y = height of printed area.

    We are told that: . xy \,=\,32 \quad\Rightarrow\quad y \,=\, \frac{32}{x}\;\;{\color{blue}[1]}

    The width of the page is: . x+2
    The height of the page is: . y+4

    The area of the page is: . A \:=\:(x+2)(y+4)

    Substitute [1]: . A \;=\;(x+2)\left(\tfrac{32}{x} + 4\right)

    And we have: . A \;=\;4x + 64x^{-1} + 40


    And that is the function we must minimize.

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