#### Answer

a. $ C(x)=180\times\sqrt {x^{2}+800^{2}}-100x+1056000$
b. The least expensive location for point Q should be less than 2000 ft

#### Work Step by Step

a. We know that 1 mile=5280ft, so 2 miles=10560ft.
The distance between the power plant and point Q is:
$ d=\sqrt {x^{2}+800^{2}}$ and the total cost is:
$ C(x)=180\times\sqrt {x^{2}+800^{2}}+100\times(10560-x)=180\times\sqrt {x^{2}+800^{2}}-100x+1056000$
b. See the table below.
As the cost $ C(x)$ increases across $ x=2000ft $,
the least expensive location for point Q should be less than 2000 ft