i'll start with a simple solution that doesn't take into account the infinite possibilities inherent in the more complex solution.
the simple solution involves a right triangle.
the more complex solution involves linear programming techniques.
basically you have a right triangle.
one leg of the right triangle is 2 miles from the house to the road.
that equals the height of your right triangle.
the other leg of the right triangle is the 5 miles from the point in the road where your house connects to it and the connection box if you travel on the road to the connection box from there.
that equals the base of the right triangle.
the height of your right triangle is 2 and the base of your right triangle is 5.
the hypotenuse of your right triangle is equal to sqrt(5^2 + 2^2) which is equal to sqrt(29).
the hypotenuse is the straight line distance from the connection box to your house and it is all off road.
the cost is therefore equal to sqrt(29) * 350 = 1885 rounded to the nearest integer.
without taking into account infinite number of possibilities in between, the other option is to go 5 miles on the road and then 2 miles off the road to reach the house.
the cost for this would be 5 * 250 + 2 * 350 = 1950.
going straight to the house from the connection box off road is cheaper than going 5 miles down the road and then 2 miles off the road to reach the same house.
this solution is valid but doesn't take into account the infinite possible combinations of x and y to get to the house.
you can travel part way down the road and then travel the rest of the way off the road to get to the same house.
the solution including these infinite number of possibilities can be solved with a linear programming type of solution.
that solution is a graphical solution since we are dealing with two dimensions only.
it is solved as follows:
x is the number of miles on the road.
y is the number of miles off the road.
your cost equation is cost = 250 * x + 350 * y.
that's your objective function that you want to minimize.
your constraints are:
x <= 5
y >= 2
y <= sqrt(29)
x >= 0
y >= 0
x^2 + y^2 = 29
the explanations for these constraints is given below:
x <= 5
the distance traveled on the road must be less than or equal to 5.
it actually could be more than 5 but that would not be smart and clearly more expensive.
y >= 2
the minimum distance from the road to the house is 2 miles.
y >= sqrt(29)
the maximum distance off the road is the straight line distance from the connection box to the house which is the hypotenuse of the right triangle.
x >= 0
can't have negative miles on the road.
y >= 0
can't have negative miles off the road.
x^2 + y^2 = 29
since the distance to the house is part on the road and part off the road and the road and the house form a right triangle, then pythagorus says that a^2 + b^2 = c^2 which is the same as x^2 + y^2 = 29 if you let a = x and b = y and c = 29.
you would grpah your constraints and then find the feasible region and then find the corner points of the feasible region and then analyze the objective function at those corner points to find the minimum cost solution.
the following graph shows the graph of the constraints.
see below the graph for additional comments.
your feasible region is within the white box.
the region that is not feasible is shaded.
i accomplished that by reversing the inequalities.
where the constraint was x >= 0, i graphed x <= 0, etc.
shading all regions that are not feasible makes it easier to see the region that is feasible.
it stands out a lot better.
the corner points of that white box are:
(0,5.385)
(0,2)
(5,5.385)
(5,2)
the minimum cost would be at one of these corner poiints.
however, there is the additional constraint that x^2 + y^2 = 29
that means that the minimum cost solution also has to lie on the line that represents the graph of that equation.
the portions of that circle within the feasible region are valid points to get from the connection box to the house.
they include some miles along the road plus some miles off the road.
the corner points of the feasible region, are still the points where the minimum cost solution lies.
looking at the graph, we see that the 2 corner points of the feasible region that lie on the circle are:
(0,5.385)
(5,2)
the other corner points of the feasible region can be ignored because they don't like on the graph of the circle equation.
we evaluate the objective function at these corner points and we get:
0 * 250 + 5.385 * 350 = 1885 rounded to the nearest dollar.
5 * 250 + 2 * 350 = 1950 dollars.
the minimum cost solution is still the straight line distance from the connection box to the house.
this solution, however, also takes into account the infinite number of possible combinations of x and y to get to the house as well.