Maximise the volume of an open triangular prism 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?
 A: 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.
A: So this is an incorrect answer to your question, and maybe I can shed some light on it.  I assume there are 4 pieces of cardboard.  The two equilateral triangles, and two rectangles are cut out.  The sum of the areas of the 4 pieces is 200 cm^2.  The area of an equilateral triangle with side length x is 
$$ A = (\sqrt3) / 4 * x^2 $$
The total area of the cardboard is 2*rectangle_area + 2*triangle_area.  The total area is 200 cm^2.  So
$$ 200 = 2*L*x + [(\sqrt3)/2]*x^2$$
Solving for L yields
$$ L = 100/x - ((\sqrt3)/4)*x $$
The volume of the prism is the area of the triangle times the length.
$$ V = (\sqrt3)/2 *x^2*L$$
Substituting our expression for L gives
$$ V =  (\sqrt3)/2 *x^2*[100/x - ((\sqrt3)/4)*x]$$
This simplifies to
$$ V = 50(\sqrt3)x - (3/8)x^3$$
so then take derivative with respect to x
$$ dV/dx = 50(\sqrt3) - x^2/8$$
Set derivative equal to 0 and solve for x.
$$x = 20\sqrt(\sqrt3)$$
And finally x= 26.3 cm.
I can plug into the expression for L above and find that L = -7.6, so I must have an error.

Regards, Matt
A: The expression for the area of the shape will be:$$\frac{\sqrt{3}}{2}x^2+2Lx=200$$And the volume will be :$$V=\frac{\sqrt{3}}{4}x^2L$$
From the first equation, $$L=\frac{400-\sqrt{3}x^2}{4x}$$
Putting it in the volume equation gives:$$V=\frac{\sqrt{3}}{16}(400x-\sqrt3x^3)$$
So $$\frac{dV}{dx}=\frac{\sqrt{3}}{16}(400-3\sqrt3x^2)$$
Setting this to zero gives the value of $x=8.7738, L=7.5983$ and $ V=253.27$
