The question
A tent is in the shape of a triangular prism.
The triangular ends are equilateral triangles.
The volume of the prism is 40 $m^3$
.
Find the length and width of the rectangular base
that give a minimum surface area for the tent.
You do not need to show that the answer is a minimum.
The equations, at least the ones I think I need.
$V$ = Volume of a triangular prism
$A$ = Surface area of a triangular prism
$V$ = 40
$V=\frac{1}{2}b\cdot h\cdot l$
$A = b\cdot h + 3b\cdot l$
?
Where I get stuck
Eliminate h by rearranging and combining equations
$$40=\frac{1}{2}b\cdot h\cdot l$$
$$80=h\cdot b\cdot l$$
$$h=\frac{80}{b\cdot l}$$
Insert into surface area equation
$$A = b\cdot \left(\frac{80}{b\cdot l}\right) + 3b\cdot l$$
Which can be simplified to
$$A = \frac{80}{l} + 3b\cdot l$$
It's at this point I have no idea where to go next. I'm fairly confident their must be a third equation, but I can't think of any that link the variables.
Any help is much appreciated.
Best Answer
From what I'm understanding how the variables are named without a diagram, since the base triangles are equilateral, $h \propto b$. So then you may instead eliminate say $b$ for $\ell$ and then minimize $A(\ell)$.