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.