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Thread: Need help with these two questions: (16x^8 y^12)^1/4 (3xy^3)

  1. #1
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    Need help with these two questions: (16x^8 y^12)^1/4 (3xy^3)

    Morning all,

    Not entirely sure which forum was best to post in, so figured i'd just put it in here. Apologies if it sin the wrong place.

    Morning all,

    So I just got an assignment back, and only went wrong on two questions woo! lol..

    But, without having the actual paper back can't see where, so was wondering if someone can explain their answers to the two following questions, and hopefully means I wont make the same mistakes in the future

    1) Evaluate the following expressions using indices
    and without using the calculator, showing all steps:

    (16x^8 y^12)^1/4 (3xy^3)

    2) From a rectangular sheet of metal measuring 120 mm by 75 mm equal squares of side x are cut from each of the corners. The remaining flaps are then folded upwards to form an open box. Show that the volume of the box is given by:

    V = 9000x 390x2 + 4x3

    Find the value of x such that the volume is a maximum:

    thanks in advance for any help you can provide
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  2. #2
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    Re: Need help with these two questions: (16x^8 y^12)^1/4 (3xy^3)

    $\left(16x^8 y^{12}\right)^{1/4}(3 x y^3)=$

    $(2 x^2 y^3)(3 x y^3)=$

    $6 x^3 y^6$

    2)

    let $a=120~mm,~b=75~mm$

    $x$ is removed from all 4 sides and folded up resulting in a box of dimensions

    $l = a-2x$

    $w = b-2x$

    $h = x$

    note that $x < \min\left(\dfrac a 2, \dfrac b 2\right)~mm$

    $V = l w h = (a-2x)(b-2x)x =a b x+ (-2 a-2 b)x^2 +4 x^3$

    plugging the numbers in gets

    $V=4 x^3-390 x^2+9000 x~mm^3$

    to maximize this we set the first derivative equal to zero and solve for $x$

    $\dfrac{dV}{dx} = 12 x^2 -780x + 9000 = 0$

    $x = 15,~50$

    and we ignore $x=50$ as we've already established that $x < \dfrac{75}{2} = 37\dfrac 1 2~mm$

    Checking the 2nd derivative at this point we see

    $\dfrac{d^2V}{dV^2} = 24x - 780$

    $\left . \dfrac{d^2V}{dV^2}\right |_{x=15} = -420 < 0$

    So $x=15~mm$ results in the maximum volume of $V=60750~mm^3$
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  3. #3
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    Re: Need help with these two questions: (16x^8 y^12)^1/4 (3xy^3)

    hrm I see this is posted in algebra so the calculus I've used isn't applicable.

    there does appear to be a way to find the maximum of a cubic function without calculus.

    See here
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  4. #4
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    Re: Need help with these two questions: (16x^8 y^12)^1/4 (3xy^3)

    question 2 is a duplicate ...

    http://mathhelpforum.com/geometry/27...e-maximum.html

    quote from that thread ...

    Do i effectively just turn it into a quadratic equation after finding the derivative?
    so, calculus is apparently OK.
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