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 Connecting the Points (Posted on 2007-02-04)
Five major buildings on a campus have coordinates A(0,0), B(0,800), C(200,1000), D(400,800), and E(400,0) (where the x and y axes are scaled in units of meters). Roads must be constructed to connect all of these buildings at a cost of \$32 per linear meter (using a standard road width).

So, for example, if the point F has coordinates (200,400) and straight roads are built between A & F, B & F, D & F, E & F, and C & D, almost 2072 meters of road would be needed to connect the buildings at a cost of \$66,294. to the nearest dollar.

Given a road construction budget of \$55,900. for this project, show how you might connect the buildings within the budget constraints.

 No Solution Yet Submitted by Dennis Rating: 4.0000 (1 votes)

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 Solution Comment 8 of 8 |
First, take the drawing and divide it along its longest axis of symmetry.  This will produce a line with equation x=200.  This line will be further defined as a line segment with the two lower points connected at the lower segment end and the three higher points connected at the upper segment end.

The lower point (x1) is calculated by maximizing the difference between the sum of the lengths of the two axis-connecting line segments and x1.  The equation for this distance is [2*sqrt(x1^2+400)-x1].  Taking the derivative and setting to zero yields [x1=200*sqrt(3)/3].  P(x1)=(200,115.47).

For the upper point (x2), the equation changes slightly to [2*sqrt(x2^2+400)+(200+x2)-x2].  Setting the derivative to zero yields x2=0.  P(x2)=(200,800).  This means that the two outlying upper points connect perpendicular to the axis.  This leaves an axis line segment of length 684.53 m.  Each of the line segments connecting the three upper points has a length of 200 m for a total of 600 m.  Each of the line segments connecting the lower two points has length [sqrt(40000+120000/9)], or 230.94 m.  This gives a grand total of 1746.41 m.  This gives a total cost of \$55,885.13.

 Posted by hoodat on 2007-02-06 17:35:20

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