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Core 1 Question differentiation; am I doing it right? Watch

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    Q) A cuboid has a volume of 8m^3. The base of the cuboid is square with sides of length x metres. The surface area of the cuboid is A m^2.

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

    I did  x \times x = x^2 and since there are two square sides multiplied this by two to get 2x^2.

    For the \frac{32}{x} I just did 4 \times \frac{8}{x} since the volume is 8 then dividing by x would give surface area and I multiplied that by four because there are four rectangular surfaces.

    So basically is that enough explanation for first one?

    (ii) Find \frac{dA}{dx}.


    That is, 4x - 32x^{-2}

    (iii) Find the value of x which gives the smallest surface area of the cuboid justifying your answer.


    Honestly I didn't know how to do this one; but check if what I did was right as a guess...

    \frac{d^2A}{dx^2} = 64x^-3 + 4

    I thought since the volume is 8m^3 then surely the smallest value of each x has to be 2? I'm not sure though - I stuck than in and got a minimum area of 12M^2.

    So can someone check if I've done part 3 right especially and also if I haven't (the most likely case ) could you please explain how you do it?
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    Ah! You're doing the June 2006 past paper! That's quite a tricky one.
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    (Original post by Student#254)
    Ah! You're doing the June 2006 past paper! That's quite a tricky one.
    Yes haha!! There is no mark scheme online so I need help! Do you know the answer?
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    (Original post by Dededex)
    Yes haha!! There is no mark scheme online so I need help! Do you know the answer?
    Yes... Give me a sec.
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    (Original post by Student#254)
    Yes... Give me a sec.
    ?
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    The last part is the part I'm most concerned about.

    Hate the applications of differentiation questions.
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    (Original post by Dededex)
    ?
    I get:

    Volume=8m^3

    h \times x^2=8

    h=\frac{8}{x^2}

    SA=2x^2+4 \times\frac{8}{x^2}x

    \therefore A=2x^2+\frac{32}{x}

    \frac{dA}{dx}=4x-\frac{32}{x^2}

    0=4x-\frac{32}{x^2}

    0=4x^2-32

    \frac{32}{4}=x^3

    8^\frac{1}{3}=x

    \therefore 2=x

    \frac{d^2A}{dx^2}=4+\frac{64}{x^  3}

    at x=2, \frac{d^2A}{dx^2}>0 \therefore (min)

    I was going to scan the solutions in, but my scanner isn't working!
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    (Original post by Student#254)
    I get:

    Volume=8m^3

    h \times x^2=8

    h=\frac{8}{x^2}

    SA=2x^2+4 \times\frac{8}{x^2}x

    \therefore A=2x^2+\frac{32}{x}

    \frac{dA}{dx}=4x-\frac{32}{x^2}

    0=4x-\frac{32}{x^2}

    0=4x^2-32

    \frac{32}{4}=x^3

    8^\frac{1}{3}=x

    \therefore 2=x

    \frac{d^2A}{dx^2}=4+\frac{64}{x^  3}

    at x=2, \frac{d^2A}{dx^2}>0 \therefore (min)

    I was going to scan the solutions in, but my scanner isn't working!
    Yeah that's kind of what I did however I didn't prove x = 2, I just assumed it since 2^3 is 8

    Man I hate that question.
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    (Original post by Dededex)
    Yeah that's kind of what I did however I didn't prove x = 2, I just assumed it since 2^3 is 8

    Man I hate that question.
    Yeah that paper has to be the hardest one I've tried so far...
 
 
 
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