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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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 Optimized | Comment 4 of 8 |

The minimum will occur when all junctures are 120 degrees.

Add points F:(253,109), G:(332,771) and H:(186,880) and connect BH, CH, HG, DG, GF, AF, EF

This gives a total length of about 1704.59 for a cost of about \$54546.88

 Posted by Jer on 2007-02-05 11:56:03

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