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Thread: calculus for areas

  1. #1
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    calculus for areas

    Hi:
    I have this question that follows. It is in a differentiation section of the book. It asks to find area and value of x. But I thought when I differentiate it is to find the gradient of a tangent to a curve. What don't I know here?

    The question is as follows:
    A piece of wire of total length 12m is cut into two pieces. One piece of wite is bent into a rectangle of sides xm and 3xm and the other is bent to form the boundary of a square. Show that the total area enclosed by the rectangle and the square is given by:



    Find the value of x if the total area enclosed is a minimum. Verify that your stationary value is a minimum.

    Thank you
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  2. #2
    Super Member Deadstar's Avatar
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    Right well lets first find perimeters.

    for the rectangle its $\displaystyle x + 3x + x + 3x = 8x$.

    So for the square it's just whatever is left which is $\displaystyle 12m-8x$.

    So area of the rectangle is width times length which is $\displaystyle 3x^2$.

    Each side of the square has length $\displaystyle \frac{12m - 8x}{4} = 3m - 2x$.

    Hence area of the square is...

    $\displaystyle (3m - 2x)^2 = 9m^2 - 12x + 4x^2$

    Hence total area is...

    $\displaystyle 4x^2 + 3x^2 - 12x + 9m^2 = 7x^2 - 12x + 9m^2$.

    To find the minimum find the derivative and set it to be 0.

    To show it is a minimum find the second derivative and check if it is greater than 0.

    EDIT: I see Soroban is replying to this thread as well. He will explain this better if you are still confused.
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  3. #3
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    ok. I have worked the question based upon Deadstar's workings and got the following:

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  4. #4
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    Hello, stealthmaths!

    A piece of wire of total length 12 m is cut into two pieces.
    One piece of wire is bent into a rectangle of sides $\displaystyle x$ m and $\displaystyle 3x$ m
    and the other is bent to form the boundary of a square.

    Show that the total area enclosed by the rectangle and the square
    is given by: . $\displaystyle A \;=\;7x^2 - 12x + 9$

    Find the value of $\displaystyle x$ if the total area enclosed is a minimum.
    Verify that your stationary value is a minimum.

    The 12-meter wire is cut into two pieces of length $\displaystyle a$ and $\displaystyle 12-a$.


    Code:
          : - - - - -  12 - - - - - :
          *---------*---------------*
          : -  a  - : - -  12-a - - :

    The left piece of wire is bent into a $\displaystyle x\text{-by-}3x\text{ rectangle.}$
    Code:
                3x
          * - - - - - *
          |           |
        x |           | x
          |           |
          * - - - - - *
               3x
    The area of this rectangle is: .$\displaystyle 3x\cdot x \:=\:\boxed{3x^2}$



    We have: .$\displaystyle x + 3x + x + 3x \:=\:a\quad\Rightarrow\quad 8x \:=\:a$

    The right piece of wire has length: .$\displaystyle 12-8x$
    It is bent into a square.
    Code:
              s
          * - - - *
          |       |
        s |       | s
          |       |
          * - - - *
              s

    We have: .$\displaystyle 4s = 12 - 8x \quad\Rightarrow\quad s \:=\:3 - 2x$

    The area of this square is: .$\displaystyle \boxed{(3-2x)^2}$


    Therefore, the total area is: .$\displaystyle A \;=\;3x^2 + (9 - 6x + 4x^2) \;=\;7x^2 - 6x + 9 $

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