Math help, You have 800 ft of fencing to make a pen for hogs. If you have a river on one side of your property, what is the dimension?

For any given dimensions a square has the largest area.  This can be proved by calculus. 

So if we divide 800 ft by 3 , then 266.66...ft is the length of on side. So three sides each of 266.666...ft plus the river bank of 266.666....ft will give a square. 

266.666 .ft x 266.666 .ft = 71,111.111.... sq.ft. is the area. 

let the length =x

breadth=y

x + 2y=800

x=800-2y

Area=xy

=(800-2y)y

=800y-2y2

For maximum area dA/dy=0

dA/dy=800-4y

800-4y=0

y=200

x=800-2(200)

=400

Dimensions are length=400 ft and breadth=200 ft

Being pigs can swim you ned the fence all the way around, and the over all size is based on the wire you have and space you have. You could make a square which would be 200 by 200, or rectangle were it is 300 by 100 feet: that is up to the space you have to put the fence!

Do you know that hogs won't go into the water and drown? This a variation on the classic "puppies in a pen problem". With fencing- or any area- the closer you get to a square, with all sides even, the more area you will have in the enclosure. 

  You have 800 feet of fencing to make a pen for hogs. 

  If you have a river on one side of your property, 

  what is the dimension of the rectangular pen 

  that maximizes the area?

  L + 2W = 800 feet

Perimeter

2w + l = 800

l = -2w + 800

Area

a = lw

a = (-2w + 800)(w)

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****ALGEBRAIC SOLUTION*****

(-2w + 800)(w) are the factors of a quadratic

for a parabola the opens down

therefor the vertex is a maximum

Find the roots

(-2w + 800)(w) = 0

Root 1                        Root 2

-2w + 800 = 0               w = 0

-2w = -800

w = 400

The vertex is the midpoint of the roots

w = (400 + 0)/2

w = 200 <–––––

l = -2w + 800

l = -2(200) + 800

l = 400 <––––––

The rectangle is

400 ft by 200 ft <––––––

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CALCULUS SOLUTION

a = (-2w + 800)(w)

a = -2w² + 800w

Take first derivative

a' = -4w + 800

Check 2nd derivative

a" = -4

2nd derivative is negative therefor critical point is a maximum

Set 1st derivative to 0 to find critical point

-4w + 800 = 0

-4w = -800

w = 200 <–––––

l = -2w + 800

l = -2(200) + 800

l = 400 <–––––

The rectangle is

400 ft by 200 ft <––––––

266.6 on each of 3 sides if access to the river is not a problem .