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    can anyone explain me the question ten part 1 and 3
    coz the answer doesn't make any sense.
    here id the link http://www.mei.org.uk/files/papers/c208ja_kf9x.pdf

    im doing a level maths and i got exam on month of may and june. in addition im doing mechanics which is similar to physics and makes me mad.

    please replay asap... thanks
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    (Original post by Rockypete1)
    can anyone explain me the question ten part 1 and 3
    coz the answer doesn't make any sense.
    here id the link http://www.mei.org.uk/files/papers/c208ja_kf9x.pdf

    im doing a level maths and i got exam on month of may and june. in addition im doing mechanics which is similar to physics and makes me mad.

    please replay asap... thanks
    10. (i) Find h in terms of x.

    The volume of a cuboid is the base area times the height:

     V = hx^2

    You also know that the volume of the cuboid is 120cm^3. Therefore:

     120 = hx^2

    Rearrange to make h the subject:

     h = \frac{120}{x^2}

    Hence show that the surface area, A cm^2, of the cuboid is given by:

     A = 2x^2 + \frac{480}{x} .


    The surface area of the cuboid is the area of the six faces added together:

     A = 4hx + 2x^2

    Substitute in h:

     A  = \left(4x \times \frac{120}{x^2}\right) + 2x^2

     = \frac{480x}{x^2} + 2x^2

     = \frac{480}{x} + 2x^2

     A = 2x^2 + \frac{480}{x}


    (ii) Find  \frac{dA}{dx} and  \frac{d^2A}{dx^2} .

     A = 2x^2 + \frac{480}{x}

     A = 2x^2 + 480x^{-1}

     \frac{dA}{dx} = 4x - 480x^{-2}

     \frac{d^2A}{dx^2} = 4 + 960x^{-3}


    (iii) Hence find the value of x which gives the minimum surface area.

    The surface area is given by  A = 2x^2 + 480x^{-1} . To find the value of x which gives the minimum surface area, you need to find the value of x which gives the lowest value for A, i.e. when  \frac{dA}{dx} = 0 and  \frac{d^2A}{dx^2} > 0 .

    Equate  \frac{dA}{dx} to and solve for x:

     4x - 480x^{-2} = 0

     4x - \frac{480}{x^2} = 0

     4x^3 - 480 = 0

     x^3 - 120 = 0

     x^3 = 120

     x = \sqrt[3]{120} cm

    Substitute into  \frac{d^2A}{dx^2} to check that it does give a minimum value i.e. greater than  0 :

     \frac{d^2A}{dx^2} = 4 + 960x^{-3}

     \frac{d^2A}{dx^2} = 4 + \frac{960}{x^3}

     \frac{d^2A}{dx^2} = 4 + \frac{960}{(\sqrt[3]{120})^3}

     \frac{d^2A}{dx^2} = 4 + \frac{960}{120}

     \frac{d^2A}{dx^2} = 4 + 8 = 12

    Therefore  x = \sqrt[3]{120} cm gives a minimum value.

    Find also the value of the surface area in this case.

    Now we substitute the value for x into the equation for the surface area, A:

     A = 2x^2 + 480x^{-1}

     x = \sqrt[3]{120}

    This can also be written as  x = (120)^{\frac{1}{3}} , which is much easier to substitute:

     A = 2 \times \left((120)^{\frac{1}{3}}\right)  ^2 + 480 \times \left((120)^{\frac{1}{3}}\right)  ^{-1}

    Use the laws of indices to simplify it:

     A = 2 \times (120)^{\frac{2}{3}} + 480 \times (120)^{-\frac{1}{3}}

     A = 48.65... + 97.31...

     A = 145.96...

    Therefore rounding to 3 significant figures gives:

     A = 146cm^2


    Phew; that was a hell of a lot faster to do with pen and paper!
    I hope that answered your queries.
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    thanks you so much for the answer.....
 
 
 
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