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Math Help - Volume of cube with part of edge removed.

  1. #1
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    Volume of cube with part of edge removed.

    The diagram below shows a solid with volume V, obtained from a cube with edge
    a > 1 when a smaller cube with edge 1/a is removed.

    sorry for the image but there is not...so imagine a solid with an edge missing...so a small solid is missing on 1 edge..

    Let x=a − 1/a

    -Find V in terms of x
    Hence or otherwise,show that the only value of a for whichV= 4x is a= 1+ √5 devided by 2..
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  2. #2
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    Hello, domenfrandolic!

    \text{The diagram below shows a solid with volume }V,
    \text{obtained from a cube with edge }a > 1
    \text{ with a smaller cube with edge }\tfrac{1}{a}\text{ removed.}

    \text{Let }x\:=\:a - \tfrac{1}{a}

    \text{(a) Find }V\text{ in terms of }x.

    \text{(b) Hence or otherwise, show that the only value of }a
    . . . . \text{ for which }V\,=\,4x\,\text{ is: }\: a\:=\: \tfrac{1+\sqrt{5}}{2}

    (a) We have: . x \:=\:a - \frac{1}{a} .[1]

    Square: . x^2 \:=\:\left(a-\frac{1}{a}\right)^2 \:=\:a^2-2+\frac{1}{a^2} \quad\Rightarrow\quad a^2 + \frac{1}{a^2} \:=\:x^2+2 .[2]

    The volume is: . V \;=\;a^3 - \left(\frac{1}{a}\right)^3 \;=\;a^3 - \frac{1}{a^3}

    Factor: . V \;=\;\left(a - \frac{1}{a}\right)\left(a^2 + 1 + \frac{1}{a^2}\right) \;=\;\underbrace{\left(a - \frac{1}{a}\right)} \left(\underbrace{a^2+\frac{1}{a^2}} + 1\right)

    Substitute [1] and [2]: . V \;=\;(x)\,(x^2+2 + 1) \quad\Rightarrow\quad \boxed{V \;=\;x(x^2+3)}



    \text{(b) If }V \,=\,4x,\text{ we have:}

    . . a^3 - \frac{1}{a^3} \;=\;4\left(a-\frac{1}{a}\right) \quad\Rightarrow\quad \left(a^3-\frac{1}{a^3}\right) - 4\left(a-\frac{1}{a}\right) \;=\;0

    Factor: . \left(a-\frac{1}{a}\right)\left(a^2 + 1 + \frac{1}{a^2}\right) - 4\left(a - \frac{1}{a}\right) \;=\;0

    Factor: . \left(a - \frac{1}{a}\right)\left(a^2 + 1 + \frac{1}{a^2} - 4\right) \;=\;0

    . . . . . . . . . \left(a - \frac{1}{a}\right)\left(a^2 - 3 +\frac{1}{a^2}\right) \;=\;0


    We have two equations to solve . . .

    a - \frac{1}{a} \:=\:0 \quad\Rightarrow\quad a^2 - 1 \:=\:0 \quad\Rightarrow\quad \rlap{///////}a \:=\:\pm1 \;\;\text{ (Recall that }a > 1)

    a^2 - 3 + \frac{1}{a^2} \;=\;0 \quad\Rightarrow\quad a^4 - 3a^2 + 1 \;=\;0
    . . \text{Quadratic Formula: }\;a^2 \:=\:\frac{3\pm\sqrt{5}}{2}\;=\;\left(\frac{1\pm \sqrt{5}}{2}\right)^2
    . . \text{Therefore: }\;a\;=\;\frac{1 + \sqrt{5}}{2}

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