An open container is to be constructed out of 200 square centimeters of cardboard. The two end pieces are equilateral triangles. The open top is a horizontal rectangle. Find the lengths of the sides of the triangle for maximum volume of the container.
So effectively we have an equilateral triangular prism of length $L$.
I first found an equation for $L$: $$L=\frac{200-\sqrt{3}x^2}{2x}$$
I then used $V = \frac{1}{2}x^2Lsin60$ to get the volume, substituting for $L$ to get the volume in terms of $x$.
Differentiating with respect to $x$, setting equal to zero, and solving for $x$ gave me value of $x=6.2$
$$\frac{dv}{dx} = 50\sqrt{0.75} – \frac{9x^2}{8} = 0$$
x=6.204
But this answer was marked wrong by my teacher. Where had I gone wrong?
Best Answer
Let's let our variable x represent the side of the equilateral triangle. In that case, the area of the equilateral triangle is $$\frac{x^2\sqrt{3}}{4}$$
Therefore, the remaining area for the cardboard to be used by the three rectangles is $$200- \Big(\frac{x^2\sqrt{3}}{4}\Big)2$$ Because we have three sides, the area taken up by a single rectangle will be $$\frac{200- \Big(\frac{x^2\sqrt{3}}{4}\Big)2}{3}$$
Because one of the dimensions of the rectangle is x, we can divide by this to find the length, giving$$\frac{200- \Big(\frac{x^2\sqrt{3}}{4}\Big)2}{3x}$$
The total formula for the volume, therefore, is $$V(x)=\frac{x^2\sqrt{3}}{4} \cdot\frac{200- \Big(\frac{x^2\sqrt{3}}{4}\Big)2}{3x}$$ $$\frac{x^2\sqrt{3}}{4} \cdot\frac{200- \Big(x^2\sqrt{3}\Big)2}{12x}$$
$$\frac{200x^2\sqrt{3}- 2\Big(x^2\sqrt{3}\Big)^2}{48x}$$
$$\frac{200x^2\sqrt{3}- 6x^4}{48x}$$
$$\frac{100x\sqrt{3}- 3x^3}{24}$$
$$-8x^3+\frac{25\sqrt{3}}{6}x$$
The first derivative is
$$V'(x)=-24x^2+\frac{25\sqrt{3}}{6}$$
The roots of this equation are approximately -0.548 and 0.548. These are the critical points. The second derivative is
$$V''(x)=-48x$$
I will let you try and figure out which one is the minimum and which is the maximum, and what the final answer is.