Posted by **Sarah** on Tuesday, March 2, 2010 at 9:40pm.

Please help solve this,

A farmer has 600m of fence and wants to enclose a rectangular field beside a river. Determine the dimensions of the fence field in which the maximum area is enclosed. (Fencing s required on only three sides: those that aren't next to the river.)

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**4**, Saturday, September 27, 2014 at 9:30pm
adf

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**Anonymous**, Saturday, March 14, 2015 at 11:02pm
Guff ffgbbnh

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**51705 MCL student**, Wednesday, March 25, 2015 at 10:43pm
A=L*W

P=L*2W three sides

600m=L*2W

L=600-2W

substitute in the area

A=(600-2W)*W

A=600W-2W^2

convert the eq. to vertex form y=a(x-h)^2+k

A=-2W^2+600W

complete the square

A=-2(W^2-300W+22500)+22500*2

A=(W-150)^45000

w=150

substitute calculate....

L=600-2w

L=600-300 =300

dimension

150x300

- simple math crackers . -
**minto**, Friday, July 17, 2015 at 6:15am
A=L*W

P=L*2W three sides

600m=L*2W

L=600-2W

substitute in the area

A=(600-2W)*W

A=600W-2W^2

convert the eq. to vertex form y=a(x-h)^2+k

A=-2W^2+600W

complete the square

A=-2(W^2-300W+22500)+22500*2

A=(W-150)^45000

w=150

substitute calculate....

L=600-2w

L=600-300 =300

dimension

150x300

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**Amara**, Monday, November 2, 2015 at 7:16pm
A family wants to fence a rectangular play area alongside the wall of their house. The wall of their house bounds one side of the play area. If they want the play area to be exactly 400ft2, what is the least amount of fencing needed to make this? Round your answer to the nearest two decimal places.

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