A farmer decides to enclose a rectangular garden, using the side of a barn as one side of the rectangle. What is the maximum area that the farmer can enclose with 100 ft. of fence? What should the dimensions of the garden be to give this area?
The max. area that the farmer can enclose with 100 ft of fence is ___sq ft.
The dimensions of the garden to give this area is 50 ft by ___ ft.
Perimeter = 2L + 2W
One of these sides is made up of the barn, so you can eliminate either one L or one W. A square ordinarily would give you the maximum area, but the side of the barn used can vary without adding to the fence used.
If one side is 50, the other two sides would be 25 to form a rectangle. From that, you can determine the area.
A farmer has a rectangular pasture with dimensions x by 2x. If he places posts 15 ft. apart, how will he represent the distance between each post on the shorter side? on the longer side?
the area of the pool is 750 sq. ft. one side is 30 feet long . how long is other side
A farmer decides to enclose a rectangular garden, using the side of a barn as one side of the rectangle. What is the maximum area that the farmer can enclose with 120 ft. of fence? What should the dimensions of the garden be to give the farmer the max area?
Can not use calculus
What are the dimensions of the largest rectangular field that can be enclosed with 100m of wire?
To find the maximum area that the farmer can enclose with 100 ft of fence, we need to consider the dimensions of the rectangle.
Let's assume the length of the rectangle is "L" and the width is "W". Since the farmer is using the side of a barn as one side of the rectangle, we only need to fence the remaining three sides.
Therefore, the perimeter of the rectangle can be written as:
Perimeter = L + 2W
According to the question, the perimeter is given as 100 ft:
100 = L + 2W
We can solve this equation to find an expression for L in terms of W:
L = 100 - 2W
Now, let's calculate the area of the rectangle:
Area = L * W
Substituting the expression for L:
Area = (100 - 2W) * W
To find the maximum area, we can take the derivative of the area function with respect to W and set it equal to zero to find the critical point.
Let's differentiate the area function:
d(Area)/dW = 100 - 4W
Setting the derivative equal to zero:
100 - 4W = 0
4W = 100
W = 25 ft
Now, we can substitute the value of W back into the expression for L to find its value:
L = 100 - 2W = 100 - 2(25) = 50 ft
Therefore, the maximum area that the farmer can enclose with 100 ft of fence is:
Area = L * W = 50 ft * 25 ft = 1250 sq ft
The dimensions of the garden to give this area are:
50 ft by 25 ft.