Question #eb797

1 Answer
Dec 11, 2016

Let us set up the following variables:

# {(w, "Width of Building (yards)"), (l, "Length of Building (yards)"), (h, "Height of Building (yards)"), (V, "Volume of the Building (cubic yards)") :} #

We want to vary the dimensions such that we maximise #V#, i.e. find a critical point of #(dV)/(dw)# that is a maximum, so we to find a V as a function of one independent variable function #V=V(w)#

Then the volume is:

# V=wlh #

The roof costs $#d# per square foot so the cost of the roof per square yard is #3*3*d=9d#. 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:

# \ \ \ \ \ D = (wl)(9d) + (2wh+2lh)(18d) + (wl)(72d) #
# :. D = 9wld + 36whd+36lhd + 72wld #
# :. D = 81wld + 36whd+36lhd #
# :. 9wl + 4wh+4lh = D/(9d)#
# :. 9wl + 4V/l+4V/w = D/(9d) # (#D# and #d# are constants)

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