Find distance on field and highway that'll minimize cost. Also, determine total cost.

Question

A cable company is to connect cable to a house. house is located 3 km from highway, and cable source is located 20km down the highway from the closest point to the house. The cost to lay the cable on the field is $1200/km, while the cost to lay the cable along the highway is $500/km. 

Answer 1

$1200 per km on the field

$500 per km on the highway

minimize cost. the house is 3 km from the highway. the cable source is 20 km down the highway from the closest point on the highway to the house.

if 20 km for the highway and 3 for the field, cost = 10,000 + 3,600 = $13,600

if directly to the house, across the field, cost = 1200x(sqr(3^2+20^2) = 1200sqr409 = $24,268

if 10 km down the highway, then to the house, cost = 5,000 + 1200sqr109 = $17,528

if 15 km down the highway then to the house, cost = 7,500 + 1200sqr34 = $14,497

if 19 km down the highway then to the house. cost =9,500 + 1200sqr10=$13,295

if 18 km down the highway then to the house, cost = 9000 + 1200sqr13= $13,327

best guess from the above is lay cable about 19 km down the highway then across the field to the house

Cost = C(x) = 500x + 1200sqr((20-x)^2+9)= 500x +1200sqr(x^2-40x +409)

take the derivative of C(x)

and set equal to 0, solve for x

C'(x) =500x + 1200sqr((x^2+9)

C'(x) = 500 +1200x/sqr(x^2+9) = 0

5=-12x/sqr(x^2+9)

5sqr(x^2+9) = -12x

x^2 +9 = 144x^2/25

9 = (144-25)x^2/25 = 119x^2/25

x^2 = 25(9)/119

x= 15/sqr119= 15sqr119/119 = about 1.375

so lay cable 20-1.375 = 18.625 km down the highway then head across the field in a beeline, as the crow flies, but underground, directly for the house

cost = 500(18.625) + 1200sqr((9+1.375^) = 9312.5 + 3960.1 = $13,272.61 = minimum cost