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.
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.
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
whoops, I dd it for 300 of fence, not 200
To find the dimensions of the rectangle that will maximize the total area of the pen, we can use the concept of optimization. Let's assume the dimensions of the rectangle are length (L) and width (W).
Step 1: Set up the equation:
The mathematical equation for the perimeter of a rectangle is given by:
Perimeter = 2L + 2W
Given that Mr. Shah has 200 meters of fencing to enclose the pen, we can write the equation as:
2L + 2W = 200
Step 2: Solve the equation for one variable:
To simplify the equation, we can divide both sides by 2:
L + W = 100
Step 3: Express one variable in terms of the other:
Let's solve the equation for W in terms of L:
W = 100 - L
Step 4: Write the equation for the area of the pen:
The area of a rectangle is given by the formula:
Area = Length × Width
Substituting the expression for W from Step 3 into the equation, we get:
Area = L × (100 - L)
Step 5: Maximize the area equation:
To maximize the area, we can take the derivative of the equation and set it equal to zero:
d(Area)/dL = 0
Taking the derivative of the equation with respect to L:
d(Area)/dL = 100 - 2L = 0
Solving for L:
100 - 2L = 0
2L = 100
L = 50
Step 6: Determine the dimensions and maximum area:
Using the value of L (50), we can find the corresponding value of W:
W = 100 - L = 100 - 50 = 50
Therefore, the dimensions of the rectangle that will maximize the total area of the pen are L = 50 and W = 50.
To calculate the maximum area, substitute the values of L and W into the area equation:
Area = L × W = 50 × 50 = 2500 square meters.
Hence, the maximum area of the pen is 2500 square meters.