Answer :
The dimensions of the largest field that can be enclosed with 136 ft of fencing are 72 ft by 64 ft. The area of the field is 4608 ft2.
The amount of fencing needed to enclose the field is equal to the sum of the lengths of all four sides. Since 136 ft of fencing is available, the sum of the lengths of all four sides must equal 136 ft.
Let x represent the length of one side and y represent the length of the other side. Then the equation for the amount of fencing needed can be written as follows:
x + x + y + y
= 136
Simplifying, we get:
2x + 2y = 136
To find the dimensions of the largest field that can be enclosed, we must maximize the area of the field. The area of the field is equal to the product of the lengths of the two sides:
A = xy
To maximize the area of the field, we must take the derivative of A with respect to x and set it equal to 0.
A' = y
Setting A' equal to 0, we get:
y = 0
Since y cannot be 0, this equation has no solution. To maximize the area of the field, we must instead take the derivative of A with respect to y and set it equal to 0.
A' = x
Setting A' equal to 0, we get:
x = 0
Since x cannot be 0, this equation has no solution. To maximize the area of the field, we must instead solve the equation 2x + 2y = 136 for x and y.
2x + 2y = 136
2x = 136 - 2y
x = 68 - y
Substituting x = 68 - y into A = xy, we get:
A = (68 - y)y
A = 68y - y2
Taking the derivative of A with respect to y and setting it equal to 0, we get:
A' = 68 - 2y
68 - 2y = 0
2y = 68
y = 34
Substituting y = 34 into x = 68 - y, we get:
x = 68 - 34
x = 34
Therefore, the dimensions of the largest field that can be enclosed with 136 ft of fencing are 34 ft by 34 ft. The area of the field is 4608 ft2.
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