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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(2): A solution | Comment 3 of 8 |
(In reply to re: A solution by Charlie)

Yep, that's how I got to 115. ^_^  I created a function, then found its minimum, which was at 200/√3. But I didn't think it was necessary to write it like that, so I rounded it.

Edited on February 4, 2007, 9:35 pm
 Posted by TamTam on 2007-02-04 15:45:51

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