Thomas' Calculus 13th Edition

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
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