Question

Mr Shah has been involved in a sheep farming business for several years and he supplies meats and milk products to the communities. He wishes to build a rectangular sheep pen with two parallel partitions using 200 meters fence. Find the dimensions of the rectangle that will maximize the total area of the pen. Hence state the maximum area of the pen.

Answer 1

as usual, maximum area is achieved in a square pen. To show this, note that if the dimensions are x and y, we have

2x+2y=200, so
y = 100-x

The area
a = xy = x(100-x) = 100x-x^2
This is just a parabola, with its vertex at x=50.

So, the pen is 50x50, or a square.

Hmm. I see I neglected to read the "two parallel partitions" part. I assume that means that the large pen is to be divided into three smaller pens. So, if the long sides have length x and the width is y, then we have

2x+4y = 200
Follow the logic above, and you will find that the maximum area is achieved when the fencing is divided equally among the lengths and widths. That is,

2x=100 and x=50
4y=100 and y=25

So the large pen is 50x25 with three sections of whatever sizes are needed by placing the interior partitions as desired.

Answer 2

if two internal petitions the 4 x needed for cross fencing + 2 L needed for length

P = 4 x + 2 L =300 or L = 150 - 2 x

A = x L = x (150 - 2 x) = 150 x - 2 x^2

if you know calculus:
dA/dx = 0 at max = 150 - 4 x
so x = 150/4 = 37.5
then
L = 150 -2*37.5 = 75
AREA = 37.5*75 = 2812.5 M^2
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IF ALGEBRA 2 THEN COMPLETE THE SQUARE

A = 150 - 2 x^2
x^2 - 75 = -.5 A

x^2 - 75 + (75/2)^2 = -.5 A + 1406.25

(x-37.5)^2 = -.5(A - 2812.5)
which we knew of course

Answer 3

whoops, I dd it for 300 of fence, not 200