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The equation for box volume can be used to express box height in terms of the side length of the base of the box. If the side length is denoted $x$, enter the function here

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Use your result from part (a) to write a function for the surface area of cardboard used in the construction of the box. Your function should be entirely in terms of the side length of the base of the box, $x$.

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To minimise the surface area of cardboard, we need the derivative of the surface area function with respect to $x$. This derivative is $\\displaystyle \\frac{dS}{dx}=$

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The minimum surface area will occur when the length of the sides of the box base is $x=$

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Finally, the height of the box must then be $h=$

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The box volume is fixed for each student. Here it lies between 0.3 and 3$m^3$.

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This question guides students through the process of determining the dimensions of a box to minimise its surface area whilst meeting a specified volume.

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A square based box is to be constructed such that its volume is exactly {volume}m$^3$. The following questions guide you through the process of determining the base side length and box height such that the minimum amount of cardboard is used to construct the box.

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The square-based box will have volume

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\$V=x^2h,\$

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where $x$ is the base side length and $h$ is the box height. Further, the surface area $S$, of the box is

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\$S=2x^2+4xh.\$

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For this particular problem, $V={volume}$, so we have

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\$\\var{volume}=x^2h \\Rightarrow h=\\frac{\\var{volume}}{x^2},\$

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providing a function for $h$ in terms of $x$. Substituting this into the surface area function we then have

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\$S(x)=2x^2+4x\\left(\\frac{\\var{volume}}{x^2}\\right)=2x^2+4\\frac{\\var{volume}}{x}.\$

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Minima and maxima of $S(x)$ occur when $dS/dx=0$. We have

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\$\\frac{dS}{dx}=4x-4\\frac{\\var{volume}}{x^2},\$

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and so we must solve

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\$0=4x-4\\frac{\\var{volume}}{x^2}.\$

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Rearranging, we find that

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\$4x=4\\frac{\\var{volume}}{x^2}\$

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\$\\Rightarrow x^3=\\var{volume}\$

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\$\\Rightarrow x=\\var{volume}^{1/3}=\\var{xsol}.\$

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Finally, substituting this value back into the $h$ equation, we see that

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\$h=\\frac{\\var{volume}}{\\var{volume}^{2/3}}=\\var{hsol}.\$

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We should confirm that this set of dimensions does in fact produce a minimum surface area (rather than a maximum). This can be carried out by noting that the 2nd derivative of $S$ with respect to $x$ is

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\$\\frac{d^2S}{dx^2}=4+8\\frac{\\var{volume}}{x^3},\$

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which is positive for all positive $x$, and hence our proposed side length will produce a minimised box surface area.

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