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A rectangular box, closed at the top, with a square base, is to have a volume of 4000 cm^ 3 . W What must be its dimensions (length, width, height ) if the box is to require the least possible material?

A rectangular box, closed at the top, with a square base, is to have a volume of 4000 cm-example-1

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Solution

Area of square base of sides x is


Area=x^2

Volume = 4000cm^3


\begin{gathered} Volume=Bh \\ B=Base\text{ }Area \\ h=height \end{gathered}

Thus,


\begin{gathered} Volume=Bh \\ 4000=x^2h \\ \\ h=(4000)/(x^2) \end{gathered}

For the box to require the least possible material, is to simply minimize the surface area of the rectangular box

The surface Area is given as


\begin{gathered} Area=2(lw+wh+lh) \\ Since,\text{ it is a square base} \\ l=x \\ w=x \\ \\ Area=2(x^2+xh+xh) \\ Area=2(x^2+2xh) \\ Area=2(x^2+2x((4000)/(x^2))) \\ \\ Area=(16000)/(x)+2x^2 \end{gathered}

Now, we differentiate


\begin{gathered} Area=(16,000)/(x)+2x^(2) \\ A=16000x^(-1)+2x^2 \\ By\text{ differentiating} \\ (dA)/(dx)=-16000x^(-2)+4x \\ \\ At\text{ minimum area, }(dA)/(dx)=0 \\ 4x=16000x^(-2) \\ x^3=4000 \\ x=10\sqrt[3]{4} \end{gathered}

Now, to find h


\begin{gathered} h=(4000)/(x^2) \\ h=\frac{4000}{100(4)^{(2)/(3)}} \\ h=4^{(1)/(3)}*10 \end{gathered}

Therefore,


\begin{gathered} Length=10\sqrt[3]{4}cm=15.874cm\text{ \lparen to three decimal places\rparen} \\ Width=10\sqrt[3]{4}cm=15.874cm\text{ \lparen to three decimal places\rparen} \\ height=15.874cm\text{ \lparen to three decimal places\rparen} \end{gathered}

User Oyinlade Demola
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