Two buildings at opposite corners of a parking lot need to be connected by cable that will be buried under ground. It costs $11/ft to lay accross parking lot and $8/ft to lay along the sides. The lot is 300ft by 360 ft rect from pt A to the opposite corner with indents of 60ft off each side. What is cheapest path?

Question

Two buildings at opposite corners of a parking lot need to be connected by cable that will be buried under ground.  It costs $11/ft to lay accross parking lot and $8/ft to lay along the sides.  The lot is 300ft by 360 ft rect from pt A to the opposite corner with indents of 60ft off each side.  What is cheapest path?

Steve · Answer

if the cable goes from A along the 360-ft side for a distance x past the indent at A's corner to y from the indent at B's corner, then the cost will be

8(60+x) + 11√((300-x)^2+(240-y)^2) + 8(60+y)

Now, I don't think you have studied multi-variable derivatives, so there must be some other relationship between x and y.