Optimization Questions?

Question

If you have any ideas about this, or if you want to solve it, by all means. Anything would be appreciated.

The owners of a small island want to bring in electricity from the mainland. The island is 85 m from a straight shoreline at the closest point. The nearest electrical connection is 250 m along the shore from that point. It costs twice as much to install cable across water as across land. What is the LEAST EXPENSIVE way to install the cable?

The other question, which I forgot to add is A 400 m track has a shape of two semi circles at the ends of a rectangle. The straight sections of the track must be at least 100 m in length, and the radius of the semi circles must be at least 20 m. Find the DIMENSIONS of the track that encloses the MAXIMUM AREA.

Answer 1

For the first part of the problem, let C be the total relative cost of the cable, with x being the distance along the shore from the closest point.

The cost of getting from the island to shore will be 2 * sqrt (85^2 + x^2) and the cost of getting along the shore from that point to the connection will be 250 - x. The total cost will therefore be

C = 2 * sqrt (85^2 + x^2) + (250 - x)

Differentiating the cost with respect to x gives

C' = 2 / (2*sqrt (85^2) + x^2) * 2x - 1
= 2*x/sqrt (85^2 + x^2) - 1

Setting the derivative on the left side to zero and solving for x gives

x = 85 / sqrt (3)

Of course, you should check that this is indeed a minimum and not a maximum, or a critical point of inflection. You can do this by checking that the second derivative is zero or by checking the algebraic sign of the derivative on either side of the candidate minimum. At x = 0, the derivative is -1, which is indeed less than zero. At x = 250, the derivative becomes nearly but not quite 1, which is positive. There are no other points besides 85/sqrt(3) within the range being considered where the derivative changes sign or becomes undefined. Therefore, the point at 85/sqrt(3) meets the criterion of being a minimum for the cost.

For the second problem,

let d be the diameter of each semicircle at the end of the track or the distance between the straight sections,
L be the length of each straight section,
and A be the enclosed area.

We then have two equations in two unknowns, d and L:

400 = pi*d + 2*L
A = pi/4*d^2 + d*L

By solving the first equation for L and substituting this expression into the equation for area, we get

A = pi/4 * d^2 + d * ((400 - pi*d)/2)
= 200*d - pi/4 * d^2

Differentiating A with respect to d gives

A' = 200 - pi/2 * d

Setting the derivative to zero and solving for d, we get

d = 400 / pi

Again we should check that this is really a maximum, but that is easy to do because the area becomes zero at d = 0. When L = 0, d also becomes 400 / pi, so the maximum area is enclosed when the track is circular and has no straight sections.

This violates the conditions of the problem because L has to be at least 100 meters. However, we have already established that the larger d is, and the smaller L is, the greater the area enclosed by the track. The dimensions of the track must therefore be as small as L can be and still mee the necessary conditions. This occurs when

L = 100,
and d = (400 - 2*100) / pi, or
d = 200 / pi

Answer 2

Let the coast line distance be X
Let the diagnal distance over water by L
Let the cost be C

L= sqrt(85^2 + (250 - x)^2)
C=X + 2L

Put the first equation into the second
Simplify
Differentiate the equation in respect to X
Find the critical number
Take the second derivative
Find the minima by finding which one of the critical number at the second derivative is negative (concave up)
That is the value for X
Then figure out the rest by plugging the value back into the original equation.

Answer 3

The total area is xy+pir^2 and you can replace the r^2 with y and the circumference of the circle is xy+2x so
400=xy+2x so y=400-2x/pi
Total area=xy + piy^2
you sub y in the total area and then you multiply them in then take the derivative and take 1/pi out so it factors better and you should get 1/pi(400-4x-1600+8x)
Then let 0=(above)
and for x you should get 300, 100, and 200-20pi and you sub it in the original and the restrictions are 400-2x/pi greater than 40 so 400-2x greater than 40pi and you get 100 lessthan x less than 200-20pi, after you have to solve for y which is around 63.66 you sub x which is 100 into 400-2x/pi I'm in calculus too

Answer 4

Write the equation for the cable cost, using perhaps the coastline distance as a variable. Then differentiate that equation with respect to the coastline distance. Look for a minimum.

Its been years since I have done this or I would give more details.

Answer 5

Hint to the first one is a slope of less then 2:1 for the angle from the island to the beach. and then I would consider figuring out the cost of putting in a solar grid.