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    https://a44694f152ee562e38a9454969d2...%20Edexcel.pdf

    help with question 7a because why not?

    Edit: Actually i need help with part b
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    It should be clear that the total length of the 12 edges is L=4(2x)+4x+4y=12x+4y where y is the dimension of the cuboid that you aren't given.

    But you know that the volume is 81, so (x)(2x)(y)=81 \Rightarrow y=\dfrac{81}{2x^2}. Therefore L=12x+4y=12x+\dfrac{162}{x^2}, as required.

    EDIT: Not sure why part (b) should cause any problems. It's your typical differentiate, set equal to 0, solve, question.
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    (Original post by IrrationalRoot)
    It should be clear that the total length of the 12 edges is L=4(2x)+4x+4y=12x+4y where y is the dimension of the cuboid that you aren't given.

    But you know that the volume is 81, so (x)(2x)(y)=81 \Rightarrow y=\dfrac{81}{2x^2}. Therefore L=12x+4y=12x+\dfrac{162}{x^2}, as required.
    Thanks for that what about part b then?
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    (Original post by thefatone)
    Thanks for that what about part b then?
    You have an expression for L in terms of x, differentiate and set it equal to 0 to find the minimum points.
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    (Original post by Zacken)
    You have an expression for L in terms of x, differentiate and set it equal to 0 to find the minimum points.
    derp, thanks so much, i'm forgetting everything, although things are coming back slowly to me ^-^
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    (Original post by thefatone)
    derp, thanks so much, i'm forgetting everything, although things are coming back slowly to me ^-^
    No problem.
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    (Original post by Zacken)
    No problem.
    where my answer was x^3=27

    then

    x=\sqrt[3] 27



x=3\sqrt3

    i don't need to put a \pm do i?
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    (Original post by thefatone)
    where my answer was x^3=27

    then

    x=\sqrt[3] 27



x=3\sqrt3

    i don't need to put a \pm do i?
    No, x^n is injective for odd n. i.e: you only use the \pm when you have x^(even number)
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    (Original post by Zacken)
    No, x^n is injective for odd n. i.e: you only use the \pm when you have x^(even number)
    i though so thanks
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    (Original post by thefatone)
    where my answer was x^3=27

    then

    x=\sqrt[3] 27



x=3\sqrt3

    i don't need to put a \pm do i?
    Do you mean x=3?
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    (Original post by thefatone)
    i though so thanks
    Awesome.
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    (Original post by Math12345)
    Do you mean x=3?
    uh oh
    i put in  \sqrt 27 into my calculator thanks

    (Original post by Zacken)
    Awesome.
 
 
 
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