# A farmer with 8000 meters of fencing wants to enclose a rectangular plot that borders on a river. If the farmer does not fence the side along the river, what is the largest area that can be enclosed?

Does that mean I have to consider it a triangle?

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1. No, make the river one of the side, fence the other three sides.

Area= LW
8000=2W+L
or L=8000-2W
Area= W(8000-2W)= 8000w-2W^2

You can find the max several ways, graphing is simple. IF you get stuck, repost.

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bobpursley
2. Thanks. I'm still confused though. I'm not sure how you got
Area= W(8000-2W)= 8000w-2W^2

What happened to the L

Do I need a system of equations?

Sorry, this has got me stumped.

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Let y= 8000x-2x^2

graph y vs X on your graphing calc, notice where the max is on x

Second method. The parabola goes up to a max then down. Find the intercepts for y=0, those will be symettrical to the parabolic axis, so look for where the midpoint of the intercepts are.
y=x(8000-2x)
intercepts x=0 , x=4000, so the max will be at x (or width 2000).
then solve for L (8000-2W).
Third method:
Calculus (in a few years you will master this, just watch now)
Area= 8000x-2x^2
d Area/dx=0= 8000-4x
solve for x, x=2000 at max.

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4. Ok, thanks so much for your explanations.

So, is the max area 8,000,000?

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5. can you help me with this homework please

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6. What don't you understand?

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