Maximum/Minimum Problem

The owners of a small island want to bring in electricity from the mainland. The island is 80m from a straight shoreline at the closest point. The nearest electrical connection is 200m along the shore from that point. It costs twice as much to install cable across water than across land. What is the least expensive way to install the cable?

The answer to this question is "lay 92.4m below water and 158.3m along the shoreline."

How do I go about solving this problem?

Thanks (:

Should be easy, but the answer is confusing, as you have cable below the water, and no mention of that cost. Not only that, but 92.4 meters below the water. Strange.

Let z be the distance from a point along the shore opposite the island to where the line crosses the shore.

Then there is a right triangle with z one side, 80 the other, and the hypotenuse, call it h, where the wire is actually laid. h² = 80² + z²
h = √(80² + z²)

the length of wire along the shore is 200–z

total cost is the partial costs added
let y be the cost per meter over land, 2y cost per meter over water
C = 2y√(80² + z²) + y(200–z)

relative cost is R = C/y. We only need relative to get min/max
R = 2√(80² + z²) + 200 – z

To get min max, differentiate and set equal to zero.
For the first part, f = 2√(6400 + z²)
we need the chain rule.
fand I run out of time, sorry.

.

Draw a right triangle whose hypotenuse is the underwater part of the cable. Call it h....

h = √(y²+80²) where y is the distance from the near point to where the cable emerges from the water.

Call the cable length along the shore Ly = 200 - y

Total cost C = 200 - y + (2)*√(y²+80²)

I set the derivative dC/y = 0 and solved for y.

y = 20.66 m → 179.34 m along the shore and 82.62 m underwater

The answer you give is inconsistent with the problem input:
If 92.4 m are underwater, y = √(92.4² - 80²) = 46.23 m → Ly = 200 - 46.23 = 153.77 m, not 158.3 m