# Question eb797

Dec 11, 2016

Let us set up the following variables:

$\left\{\begin{matrix}w & \text{Width of Building (yards)" \\ l & "Length of Building (yards)" \\ h & "Height of Building (yards)" \\ V & "Volume of the Building (cubic yards)}\end{matrix}\right.$

We want to vary the dimensions such that we maximise $V$, i.e. find a critical point of $\frac{\mathrm{dV}}{\mathrm{dw}}$ that is a maximum, so we to find a V as a function of one independent variable function $V = V \left(w\right)$

Then the volume is:

$V = w l h$

The roof costs $$d$per square foot so the cost of the roof per square yard is $3 \cdot 3 \cdot d = 9 d$. Therefore the material costs of the building, and the Surface Areas are given by:  {: ( "Component", "Surface Area sq yards", "$ cost per sq yard" ), ( "Foundation", wl, 4*("walls")=72d), ( "Walls", 2wh+2lh, 2*("roof")=18d), ( "Roof",wl, 9d) :} #

So the Total material cost, $D$, is given by:

$\setminus \setminus \setminus \setminus \setminus D = \left(w l\right) \left(9 d\right) + \left(2 w h + 2 l h\right) \left(18 d\right) + \left(w l\right) \left(72 d\right)$
$\therefore D = 9 w l d + 36 w h d + 36 l h d + 72 w l d$
$\therefore D = 81 w l d + 36 w h d + 36 l h d$
$\therefore 9 w l + 4 w h + 4 l h = \frac{D}{9 d}$
$\therefore 9 w l + 4 \frac{V}{l} + 4 \frac{V}{w} = \frac{D}{9 d}$ ($D$ and $d$ are constants)

And so we have reduced the problem to the Volume being a function of two variables $V = V \left(w , l\right)$. I could go on and look at the partial derivatives, $\frac{\partial V}{\partial l}$ and $\frac{\partial V}{\partial w}$ but I suspect the question is not that "deep" and instead another constraint is missing.