+97 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.
+118690
$$Let the triangle have side length 2a, the perpendicular height will be $a\sqrt3$\\
Let the depth of prism be D (all units in cm)\\
A $1cm^3$ container has the capacity of $1mL$\\
Volume of prism = Area of triangle $\times$ Depth\\
$5178 = \frac{1}{2} \times 2a \times a\sqrt3 \times D$\\
$5178 = a^2\sqrt3 \times D$\\\\
$D=\dfrac{5178}{a^2\sqrt3} $\\$$
$$$D=\dfrac{5178}{a^2\sqrt3} $\\\\
Surface area\\
S=2(triangles)+3(rectangles)\\\\
$S=2(\frac{1}{2}\times 2a \times a\sqrt3)+3(2a\times D)$\\\\
$S=2(a^2\sqrt3)+3(2a\times \frac{5178}{a^2\sqrt3})$\\\\
$S=2\sqrt3a^2+\frac{5178*6}{a\sqrt3}$\\\\
$S=2\sqrt3a^2+\frac{31068a^{-1}}{\sqrt3}$\\\\$$
$$Surface area will be minimum when \\\\
$\frac{dS}{da}=0\;\;and\;\;\frac{d^2S}{da^2}>0$\\\\
$\frac{dS}{da}=4\sqrt3a-\frac{31068a^{-2}}{\sqrt3}$\\\\
$0=4\sqrt3a-\frac{31068a^{-2}}{\sqrt3}$\\\\
$0=12a-31068a^{-2}$\\\\
$0=12a^3-31068$\\\\
$12a^3=31068$\\\\
$a^3=2589$\\\\
$a\approx 13.73$cm\\\\\\
$\frac{d^2S}{da^2}=4\sqrt3+\frac{2*31068a^{-3}}{\sqrt3}>0\;for\; all \;a>0$\\\\
Therefore any turning point must be a minimum.$$
$$\\a\approx 13.73cm\\\\
D\approx \frac{5178}{13.73^2*\sqrt3}\\\\
D\approx 15.86cm\\\\\\
\mbox{The triangle has side length 2*13.73= 27.46cm and the prism is 15.86cm deep}\\\\
check\\
V=1/2*27.46*13.73*\sqrt3*15.86 = 5178.5cm^3 \qquad good$$
I think that is okay 
+118690
$$Let the triangle have side length 2a, the perpendicular height will be $a\sqrt3$\\
Let the depth of prism be D (all units in cm)\\
A $1cm^3$ container has the capacity of $1mL$\\
Volume of prism = Area of triangle $\times$ Depth\\
$5178 = \frac{1}{2} \times 2a \times a\sqrt3 \times D$\\
$5178 = a^2\sqrt3 \times D$\\\\
$D=\dfrac{5178}{a^2\sqrt3} $\\$$
$$$D=\dfrac{5178}{a^2\sqrt3} $\\\\
Surface area\\
S=2(triangles)+3(rectangles)\\\\
$S=2(\frac{1}{2}\times 2a \times a\sqrt3)+3(2a\times D)$\\\\
$S=2(a^2\sqrt3)+3(2a\times \frac{5178}{a^2\sqrt3})$\\\\
$S=2\sqrt3a^2+\frac{5178*6}{a\sqrt3}$\\\\
$S=2\sqrt3a^2+\frac{31068a^{-1}}{\sqrt3}$\\\\$$
$$Surface area will be minimum when \\\\
$\frac{dS}{da}=0\;\;and\;\;\frac{d^2S}{da^2}>0$\\\\
$\frac{dS}{da}=4\sqrt3a-\frac{31068a^{-2}}{\sqrt3}$\\\\
$0=4\sqrt3a-\frac{31068a^{-2}}{\sqrt3}$\\\\
$0=12a-31068a^{-2}$\\\\
$0=12a^3-31068$\\\\
$12a^3=31068$\\\\
$a^3=2589$\\\\
$a\approx 13.73$cm\\\\\\
$\frac{d^2S}{da^2}=4\sqrt3+\frac{2*31068a^{-3}}{\sqrt3}>0\;for\; all \;a>0$\\\\
Therefore any turning point must be a minimum.$$
$$\\a\approx 13.73cm\\\\
D\approx \frac{5178}{13.73^2*\sqrt3}\\\\
D\approx 15.86cm\\\\\\
\mbox{The triangle has side length 2*13.73= 27.46cm and the prism is 15.86cm deep}\\\\
check\\
V=1/2*27.46*13.73*\sqrt3*15.86 = 5178.5cm^3 \qquad good$$
I think that is okay
