Question #b9993

1 Answer
Jan 30, 2017

The total length of the partition should be #125# feet, and the height of each partition should be #50# feet (making each section #50# feet x #31.25# feet). This results in a maximum enclosed area of #6250 " feet"^2 #

Explanation:

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Without loss of generality let us assume that the pen is divided as shown:

Let us set up the following variables:

# {(x, "Total height of the partition (feet)"), (y, "Total length of the partition (feet)"), (A, "Total Area enclosed by the partition (sq feet)") :} #

Our aim is to find #A(x)#, and to maximize the total area, A, wrt x (equally we could the same with #y# and we would get the same result). ie we want a critical point of #(dA)/dx#.

Now, the total perimeter is given as #500# (constant) and so:

# 5x + 2y=500 #
# :. 2y=500 - 5x#
# :. y=250 - 5/2x# ..... [1]

And the total Area enclosed by the pen is given by:

# A =xy #

And substitution of the first result [1] gives us:

# A =x(250 - 5/2x) #
# \ \ \ = 250x - 5/2x^2 #

We no have the Area, A, as a function of a single variable, so Differentiating wrt #x# we get:

# (dA)/dx = 250 -5x # ..... [2]

At a critical point we have #(dA)/dx=0 => #

# 250-5x = 0 #
# :. \ \ \ \ \ 5x = 250 #
# :. \ \ \ \ \ \ \ x = 50 #

And substituting #x=50# into [1] we get;

# y=250 - 5/2(50)#
# \ \ =250 - 125#
# \ \ =125#

We should check that #x=50# results in a maximum area. Differentiating [2] wrt x we get:

# (d^2A)/dx^2 = -5 < 0 # when #x=50#

Confirming that we have a maximum area, given by:

# A = (50)(125) = 6250 " feet"^2 #

We can visually verify that this corresponds to a maximum by looking at the graph of #y=A(x)#:
graph{250x - 5/2x^2 [-100, 200, -100, 7000]}