In: Advanced Math
The depth of patio is and of pool is
. So,
their respective areas are
and
. The patio area
at least as large as pool area says
. The cost
for pool area is
and that of patio is
.
Pool area is at least
sq ft, so
. Finally,
.
Total cost is
. Total backyard area unused is
.
Thus, the LP is
Simplifying the constraints gives
We solve this using graphical method. The graph of the constraints
is as follows:
Thus, the corner points are
,
,
, and
. At these points, the value of the objective function,
is
,
,
, and
. Thus, the minimum is at
, and it is
.
Thus, the optimal solution is:
,
. Patio is
ft deep, pool is
ft deep.