What is the largest possible area of 5 pens with 600ft of fencing?

A rancher wants to enclose a rectangular area and then divide it into 5 pens with fencing parallel to one side of the rectangle. He has 600 feet of fencing available to complete the job. What is the largest possible total area of the five pens?





How do you set up a problem like this to find area?|||consider the entire figure:


You have a rectangle with 4 lines going though the middle (to subdivide it into 5 pens). Let x be the side parallel to the middle fences, and let y be the other side. Note that, by the confines of the problem,


2y + 2x + 4x = 600, since you have 600 ft of fencing. This simplifies to:


2y + 6x = 600


y = 300 - 3x





This is useful because now you can write the total area of the rectangle as


A = xy = x(300 - 3x) = 300x - 3x^2


Now, finding the maximum is just a matter of differentiating and finding the critical points:





A' = 300 - 6x = 0


6x = 300


x = 50





Thus, your maximum area occurs when you choose a length x of 50. Your total area is:


A = x(300 - 3x) = 50(150) = 7500 square feet


Thank you for the correction Siress :-)





good luck ranching!|||Picture this:


_____


| | | | | |


| | | | | | y


| | | | | |


---------


x





So:


6y + 2x = 600 (feet of fencing)


y = 100 - x/3





Area:


f(x)=x*y = 100x -(x^2)/3


[this is the easy method, requires you to take a derivative]


f'(x) = 100 - 2*x/3


f'(x) = 0 at max and min, since this is 2ed degree, only has one or the other; since x^2 is negative, it's a maximum.


100 - 2*x/3 = 0


x= 150


y= 100 - 150/3 = 50





Self check:


6(50) + 2(150) ?= 600 ....check.





Then, the maximum area is 150*50 = 7500








EDIT: Ben, you need to make a correction:


"A = x(300 - x) = 50(250) = 12500 square feet"


A = x(300 - 3*x) = 7500|||5 squares, so divide 600 by 5, 120


120 ft per pen


divided by 4, 30x30 pens


30x30=900


900*5=4500sqr feet