SOLUTION: A piece of wire 57 inches long is cut into two pieces, and two rectangles are formed from the pieces. If you want the sum of the areas of the two rectangles to be as large as possi

Question 70112: A piece of wire 57 inches long is cut into two pieces, and two rectangles are formed from the pieces. If you want the sum of the areas of the two rectangles to be as large as possible, how long should you make the pieces?
Answer by checkley75(3666) (Show Source): You can put this solution on YOUR website!
THE LARGEST AREAS OF A TRIANGLE IS A SQUARE THUS TO FORM 2 SQUARES.
ONE SQUARE NEEDS TO HAVE A PERIMETER OF 56.999999999 INCHES THUS THE AREA IS (56.999999999/2)^2
(28.499999999)^2
812.2499 SQUARE INCHES BY THE AREA OF THE .000000001 PIECE OF WIRE.
OF COURSE IF YOU DON'T CUT THE WIRE YOU CAN GET
(57/2)^2=812.25 IN^2
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CUTTING THE WIRE IN HALF IS THE WORST SOLUTION THUS
57/2=28.5 FOR BOTH PIECES OF WIRE.
EACH WIRE WILL ENCLOSE AN AREA OF
(28.5/2)^2
203.0625 IN^2 FOR ONE PIECE AND 203.0625 FOR THE OTHER PIECE WHICH TOTALS ONLY
203.0625*2=406.125 IN ^2

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