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.)

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

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

- Calc. -
**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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