A farmer with 3000 feet wants to enclose a rectangular plot that borders on a straight highway. If the farmer does not fence the side along the highway, what is the largest area that can be enclosed?
Let x be the side length parallel to the highway. The side lengths perpendicular to the highway must be
(1/2) (3000 -x)
The area is
A = (x/2)*(3000-x) = 3000x -x^2/2
When there is a maximum,
dA/dx = 0 = 3000 - 2x
x = 1500 feet
I used calculus but you can try completeing the square or try various values of x until you get a maximum area.
The enclosed area will be 1500 x 750 = 1,125,000 ft^2
where did you get the 750? I have a similar problem here:
a farmer wants to build a rectangular pen using a side of a barn and 60ft of fence. find the dimensions and area of the largest such pen
Let l = measure of the parallel side of the highway in meters
w = measure of the perpendicular side of the highway in meters
l + w = 3000
l = 3000 - w
length = 3000 - w
width = w
A= lw
**since we will only use 1 side of the length, we will use:
A= [(3000-w)/2]w 0r w[(3000-w)/2]
= -(w^2)/2 + 1500w
**complete the square
= -1/2 (w^2 - 3000w + 225,000) + 1,125,000
= -1/2 (w-1500)^2 + 1,125,000
w=1500
A max.= 1,125,000
**substitution
l=(3000-w)/2
=(3000-1500)/2
=750
dimensions: 1500 x 750
To find the largest area that can be enclosed, we need to understand the relationship between the dimensions of the rectangular plot and its area.
Let's assume the length of the rectangular plot is L and the width is W.
The perimeter of the plot is given as 3000 feet. For a rectangle, the perimeter is calculated by adding the lengths of all four sides:
Perimeter = 2L + 2W
Since the farmer does not fence the side along the highway, one of the sides will not contribute to the perimeter. In this case, we can ignore one side, which means the perimeter equation becomes:
3000 = L + 2W
We can rearrange this equation to express L in terms of W:
L = 3000 - 2W
Now we have an equation for the length in terms of the width.
The area of a rectangle is calculated by multiplying the length by the width:
Area = L * W
Substituting the value of L from the rearranged equation, we get:
Area = (3000 - 2W) * W
To find the largest area that can be enclosed, we need to find the maximum value of this area equation.
We can do this by taking the derivative of the area equation, setting it equal to zero, and solving for W. This will give us the width that maximizes the area.
To find the derivative, let's use the product rule:
Area' = (3000 - 4W) * dW/dW + W * d(3000 - 2W)/dW
= (3000 - 4W) + W * (-2)
= 3000 - 4W - 2W
= 3000 - 6W
Setting the derivative equal to zero:
3000 - 6W = 0
6W = 3000
W = 500
Now we have the value of the width, W, that maximizes the area. We can substitute this value back into the area equation to find the maximum area:
Area = (3000 - 2(500)) * 500
Area = (3000 - 1000) * 500
Area = 2000 * 500
Area = 1,000,000 square feet
Therefore, the largest area that can be enclosed with a perimeter of 3000 feet, when one side is not fenced, is 1,000,000 square feet.