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The length of the box is \\(\\var{l}-2x\\), the width is \\(\\var{w}-2x\\) and the height is \\(x\\).
\nThe volume is then given by
\n\\(V=(\\var{l}-2x).(\\var{w}-2x).x\\)
\n\\(V=\\simplify{4x^3-2*({w}+{l})x^2+{l}*{w}x}\\)
\nTo find the maximum volume, we need to find $x$ such that $\\frac{dV}{dx} =0$, i.e.
\nSolve \\(\\frac{dV}{dx}=\\simplify{12x^2-4*({w}+{l})x+{l}*{w}}=0\\)
\nThis is a quadratic equation.
\n\\(x=\\frac{\\simplify{4*({w}+{l})}\\pm\\sqrt(\\simplify{16*({w}+{l})^2-48*{w}*{l}})}{24}\\)
\n\\(x=\\frac{\\simplify{{w}+{l}}\\pm\\sqrt(\\simplify{{w}^2-{w}*{l}+{l}^2})}{6}\\)
\n\\(\\frac{d^2V}{dx^2}=\\simplify{24x-4*({w}+{l})}\\)
\nwhen \\(x=\\simplify{({w}+{l}-sqrt({w}^2-{w}*{l}+{l}^2))/6}\\) \\(\\frac{d^2V}{dx^2}<0\\) and therefore is the value that gives a maximum.
", "statement": "A rectangular sheet of metal of length = \\(\\var{l}cm\\) and width = \\(\\var{w}cm\\) has a square of side \\(x\\,cm\\) cut from each corner. The ends and sides will be folded upwards to form an open box.
\nDetermine the value of \\(x\\) that will maximise the volume of this box.
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\n\\(x = \\) [[0]]
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