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# Working out the triangle length and prism length of a equilateral triangular prism?

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A chocolate company needs to manufacture cardboard containers with the shape of equilateral-triangular prisms and that have a volume of exactly 5178 mL. What are the optimum dimensions (in cm) such that a minimum amount of cardboard is used?

Triangle edges =

Prism length =

This is how the question is written.

Can you please show working out, thanks again.

Oli96  Aug 28, 2014

#1
+92217
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$$Let the triangle have side length 2a, the perpendicular height will be a\sqrt3\\ Let the depth of prism be D (all units in cm)\\ A 1cm^3 container has the capacity of 1mL\\ Volume of prism = Area of triangle \times Depth\\ 5178 = \frac{1}{2} \times 2a \times a\sqrt3 \times D\\ 5178 = a^2\sqrt3 \times D\\\\ D=\dfrac{5178}{a^2\sqrt3} \\$$

$$D=\dfrac{5178}{a^2\sqrt3} \\\\ Surface area\\ S=2(triangles)+3(rectangles)\\\\ S=2(\frac{1}{2}\times 2a \times a\sqrt3)+3(2a\times D)\\\\ S=2(a^2\sqrt3)+3(2a\times \frac{5178}{a^2\sqrt3})\\\\ S=2\sqrt3a^2+\frac{5178*6}{a\sqrt3}\\\\ S=2\sqrt3a^2+\frac{31068a^{-1}}{\sqrt3}\\\\$$

$$Surface area will be minimum when \\\\ \frac{dS}{da}=0\;\;and\;\;\frac{d^2S}{da^2}>0\\\\ \frac{dS}{da}=4\sqrt3a-\frac{31068a^{-2}}{\sqrt3}\\\\ 0=4\sqrt3a-\frac{31068a^{-2}}{\sqrt3}\\\\ 0=12a-31068a^{-2}\\\\ 0=12a^3-31068\\\\ 12a^3=31068\\\\ a^3=2589\\\\ a\approx 13.73cm\\\\\\ \frac{d^2S}{da^2}=4\sqrt3+\frac{2*31068a^{-3}}{\sqrt3}>0\;for\; all \;a>0\\\\ Therefore any turning point must be a minimum.$$

$$\\a\approx 13.73cm\\\\ D\approx \frac{5178}{13.73^2*\sqrt3}\\\\ D\approx 15.86cm\\\\\\ \mbox{The triangle has side length 2*13.73= 27.46cm and the prism is 15.86cm deep}\\\\ check\\ V=1/2*27.46*13.73*\sqrt3*15.86 = 5178.5cm^3 \qquad good$$

I think that is okay

Melody  Aug 28, 2014
Sort:

#1
+92217
+10

$$Let the triangle have side length 2a, the perpendicular height will be a\sqrt3\\ Let the depth of prism be D (all units in cm)\\ A 1cm^3 container has the capacity of 1mL\\ Volume of prism = Area of triangle \times Depth\\ 5178 = \frac{1}{2} \times 2a \times a\sqrt3 \times D\\ 5178 = a^2\sqrt3 \times D\\\\ D=\dfrac{5178}{a^2\sqrt3} \\$$

$$D=\dfrac{5178}{a^2\sqrt3} \\\\ Surface area\\ S=2(triangles)+3(rectangles)\\\\ S=2(\frac{1}{2}\times 2a \times a\sqrt3)+3(2a\times D)\\\\ S=2(a^2\sqrt3)+3(2a\times \frac{5178}{a^2\sqrt3})\\\\ S=2\sqrt3a^2+\frac{5178*6}{a\sqrt3}\\\\ S=2\sqrt3a^2+\frac{31068a^{-1}}{\sqrt3}\\\\$$

$$Surface area will be minimum when \\\\ \frac{dS}{da}=0\;\;and\;\;\frac{d^2S}{da^2}>0\\\\ \frac{dS}{da}=4\sqrt3a-\frac{31068a^{-2}}{\sqrt3}\\\\ 0=4\sqrt3a-\frac{31068a^{-2}}{\sqrt3}\\\\ 0=12a-31068a^{-2}\\\\ 0=12a^3-31068\\\\ 12a^3=31068\\\\ a^3=2589\\\\ a\approx 13.73cm\\\\\\ \frac{d^2S}{da^2}=4\sqrt3+\frac{2*31068a^{-3}}{\sqrt3}>0\;for\; all \;a>0\\\\ Therefore any turning point must be a minimum.$$

$$\\a\approx 13.73cm\\\\ D\approx \frac{5178}{13.73^2*\sqrt3}\\\\ D\approx 15.86cm\\\\\\ \mbox{The triangle has side length 2*13.73= 27.46cm and the prism is 15.86cm deep}\\\\ check\\ V=1/2*27.46*13.73*\sqrt3*15.86 = 5178.5cm^3 \qquad good$$

I think that is okay

Melody  Aug 28, 2014

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