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Math Help - Calculus and shapes crossover question! Help!!

  1. #1
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    Calculus and shapes crossover question! Help!!

    I found this one pretty hard..

    A cuboid has volume of 8 m^3 . The base of the cuboid is square with sides of length x // m . The Surface area of the cuboid is A // m^2

    1) Show that A=2x^2+\frac{32}{x}

    2) Find \frac{dA}{dx}

    3) Find the value of x which gives the smallest Surface Area of the cuboid, justifying your answer.


    Please help! I have a real exam on this soon and the rest of Core 1 is easy except these stupid little questions..

    Thanks guys!
    Last edited by anthmoo; December 27th 2006 at 12:28 PM.
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  2. #2
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    Is something wrong with LaTeX?

    Can you guys see it ok?
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  3. #3
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    [QUOTE=anthmoo;32716]I found this one pretty hard..

    A cuboid has volume of 8 m^3 . The base of the cuboid is square with sides of length x // m . The Surface area of the cuboid is A // m^2

    1) Show that A=2x^2+\frac{32}{x}[quote]

    Apparently, LaTex is down. Has been for a few days now.

    Anyway. The volume of your cuboid could be represented as:

    (x^2)y=8

    y=8/x^2

    A=2x^2+4x(8/x^2)=2x^2+(32/x)


    Now, do that differentiate thing to find parts 2 and 3
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  4. #4
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    [QUOTE=galactus;32718][QUOTE=anthmoo;32716]I found this one pretty hard..

    A cuboid has volume of 8 m^3 . The base of the cuboid is square with sides of length x // m . The Surface area of the cuboid is A // m^2

    1) Show that A=2x^2+\frac{32}{x}

    Apparently, LaTex is down. Has been for a few days now.

    Anyway. The volume of your cuboid could be represented as:

    (x^2)y=8

    y=8/x^2

    A=2x^2+4x(8/x^2)=2x^2+(32/x)


    Now, do that differentiate thing to find parts 2 and 3
    Thanks!

    I've differentiated it to 4x-32x^-2

    but how do i find the smallest value for x to give the smalles surface area?
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  5. #5
    Grand Panjandrum
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    LaTeX is not working so I will translate this into ASCII

    I found this one pretty hard..

    A cuboid has volume of 8 m^3. The base of the cuboid is square with sides of length x m . The Surface area of the cuboid is A m^2

    1) Show that A=2x^2+32/x[/tex]

    2) Find dA/dx

    3) Find the value of x which gives the smallest Surface Area of the cuboid, justifying your answer.


    Please help! I have a real exam on this soon and the rest of Core 1 is easy except these stupid little questions..

    Thanks guys!
    RonL
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  6. #6
    Grand Panjandrum
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    [quote=anthmoo;32719][quote=galactus;32718]
    Quote Originally Posted by anthmoo View Post
    I found this one pretty hard..

    A cuboid has volume of 8 m^3 . The base of the cuboid is square with sides of length x // m . The Surface area of the cuboid is A // m^2

    1) Show that A=2x^2+\frac{32}{x}

    Thanks!

    I've differentiated it to 4x-32x^-2

    but how do i find the smallest value for x to give the smalles surface area?
    The minimum surface area corresponds to the x which is a root of:

    dA/dx=0,

    so if your work is right you need to find the roots of:

    4x-32x^-2=0,

    and if there are more than one identify which corresponds to a minimum.

    A root of dA/dx=0 corresponds to a minimum of A(x) if d^2A/dx^2<0 at the root.

    RonL
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  7. #7
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    Hello, anthmoo!

    Let me walk through this one . . .


    A cuboid has volume of 8 m.
    The base of the cuboid is square with sides of length x m.
    The surface area of the cuboid is A m.

    (1) Show that: .A .= .2x + (32/x)

    (2) Find: .dA/dx

    (3) Find the value of x which gives the smallest Surface Area of the cuboid.
    Code:
                  x
              * - - - *
           x/       / |
          * - - - *   |
          |       |   |y
          |       |   |
         y|      y|   |
          |       |   *
          |       | /x
          * - - - *
              x

    We are given the volume: .V .= .xy .= .8 . . y .= .8/x . [1]


    The top and bottom have an area of: 2x
    The four sides have an area of: 4xy

    The total surface area is: .A .= .2x + 4xy . [2]

    Substitute [1] into [2]: . A .= .2x + 4x(8/x)

    Hence, we have: . A .= .2x + (32/x) . (1)

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    We have: .A .= .2x + 32x^{-1}

    Then: .dA/dx .= .4x - 32x^{-2} . (2)

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    We have: . 4x - 32x^{-2} .= .0

    Multiply by x: . 4x - 32 .= .0

    And we have: . 4x = 32 . . x = 8 . . x = 2 .(3)

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Substitute into [1]: . y .= .8/2 . . y = 2

    As you may have suspected,
    . . the cuboid with minimum surface area is a cube.

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  8. #8
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    Quote Originally Posted by Soroban View Post
    Hello, anthmoo!

    Let me walk through this one . . .


    Code:
                  x
              * - - - *
           x/       / |
          * - - - *   |
          |       |   |y
          |       |   |
         y|      y|   |
          |       |   *
          |       | /x
          * - - - *
              x

    We are given the volume: .V .= .xy .= .8 . . y .= .8/x . [1]


    The top and bottom have an area of: 2x
    The four sides have an area of: 4xy

    The total surface area is: .A .= .2x + 4xy . [2]

    Substitute [1] into [2]: . A .= .2x + 4x(8/x)

    Hence, we have: . A .= .2x + (32/x) . (1)

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    We have: .A .= .2x + 32x^{-1}

    Then: .dA/dx .= .4x - 32x^{-2} . (2)

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    We have: . 4x - 32x^{-2} .= .0

    Multiply by x: . 4x - 32 .= .0

    And we have: . 4x = 32 . . x = 8 . . x = 2 .(3)

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Substitute into [1]: . y .= .8/2 . . y = 2

    As you may have suspected,
    . . the cuboid with minimum surface area is a cube.

    Thankyou very much soroban! Hopefully now I'd be a little more prepared because you walking me through that question will give me more confidence for questions like these.

    By the way, what do you use to post? ie. the ASCII images?


    Thanks again man
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  9. #9
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    Hello again, anthmoo!

    I use these codes (when LaTeX is not working).

    Precede the number with &# and follow them with ; (semicolon)

    \begin{array}{ccccccccc} 8776 & \approx & \\ 8800 & \neq & \\ 8730 & \sqrt{} & \\ 177 & \pm & \\ 247 & \div & \\ 952 & \theta & \\ 176 & ^o &\text{degree} \\ 183 & \cdot & \text{dot} \\ 960 & \pi &  <br />
\end{array} . . . \begin{array}{ccccccccc}178 & ^2 & \text{squared} \\ 179 & ^3 & \text{cubed} \\ 8594 & \rightarrow & \\ 8592 & \leftarrow & \\ 8595 & \downarrow & \\ 8595 & \downarrow & \\ 8747 & \int & \\ 8721 &  \sum & \\ 8710 & \Delta &\end{array}


    LaTeX is back!

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