Posted by **muffy** on Saturday, September 12, 2009 at 12:22pm.

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?

- Math -
**bobpursley**, Saturday, September 12, 2009 at 12:30pm
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.

- Math -
**muffy**, Saturday, September 12, 2009 at 12:47pm
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.

- Math -
**bobpursley**, Saturday, September 12, 2009 at 12:52pm
It is a quadratic.

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.

- Math -
**muffy**, Saturday, September 12, 2009 at 1:07pm
Ok, thanks so much for your explanations.

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

- Math -
**princess**, Thursday, January 19, 2012 at 4:03pm
can you hepl me with this homework please

- Math -
**mohjee**, Friday, January 3, 2014 at 2:34pm
What don't you understand?

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