A farmer has 160 feet of fencing to enclose 2 adjacent rectangular pig pens. What dimensions should be used so that the enclosed area will be a maximum?

1 Answer
Jul 19, 2016

I'm assuming that the pig pens have identical dimensions.

Explanation:

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Let's assume that the pig pens need to be fenced in the way shown in the diagram above.

Then, the perimeter is given by #4x + 3y = 160#.

#4x = 160 - 3y#

#x = 40 - 3/4y#

The area of a rectangle is given by #A = L xx W#, however here we have two rectangles put together, so the total area will be given by #A = 2 xx L xx W#.

#A = 2(40 - 3/4y)y#

#A = 80y - 3/2y^2#

Now, let's differentiate this function, with respect to y, to find any critical points on the graph.

#A'(y) = 80 - 3y#

Setting to 0:

#0 = 80 - 3y#

#-80 = -3y#

#80/3 = y#

#x = 40 - 3/4 xx 80/3#

#x = 40 - 20#

#x = 20#

Hence, the dimensions that will give the maximum area are #20# by #26 2/3# feet.

A graphical check of the initial function shows that the vertex is at #(26 2/3, 1066 2/3)#, which represents one of the dimensions that will give the maximum area and the maximum area, respectively.

Hopefully this helps!