A farmer has 1000 feet of fencing materials available to fence a rectangular pasture next to a river. If the side along the river does not need to be fenced, what dimensions maximize the enclosed area? What is the maximum enclosed area?
let side parallel to river be y
let the other two sides be x each
2x + y = 1000 , ----> y = 1000-2x
area = xy = x(1000-2x) = 1000x - 2x^2
if you know Calculus.....
d(area)/dx = 1000 - 4x
= 0 for a max of area
4x =1000
x = 250
y = 1000-2(250) = 500
so max area = xy = 250(500) = 125000 ft^2
To find the dimensions that maximize the enclosed area, we need to use the given information about the length of fencing material available.
Let's assume the length of the side parallel to the river is x feet. Since the side along the river doesn't need to be fenced, we only need to account for the other three sides.
The total length of the other three sides would then be 1000 - x feet, since the total fencing material length is 1000 feet.
To calculate the enclosed area in terms of x, we multiply the dimensions of the length and width:
Area = length × width
Since the length is x feet, and the width is (1000 - 2x)/2, we can substitute these values into the equation:
Area = x × (1000 - 2x)/2
Simplifying this equation, we get:
Area = (1000x - 2x^2)/2
Area = 500x - x^2/2
To maximize the area, we need to find the value of x that gives us the maximum value for the area.
To do this, we can find the derivative of the area function with respect to x and then set it to zero, to find the critical points. We can then check which critical point corresponds to the maximum.
Differentiating the area function, we get:
d(Area)/dx = 500 - x/2
Setting the derivative equal to zero:
500 - x/2 = 0
Solving for x, we find:
x = 1000/2
x = 500
So, the critical point is x = 500.
To confirm this point corresponds to the maximum, we can find the second derivative of the area function and evaluate it at x = 500.
Taking the second derivative, we get:
d^2(Area)/dx^2 = -1/2
Since the second derivative is negative, this confirms that x = 500 corresponds to the maximum.
Therefore, the dimensions that maximize the enclosed area are:
Length = x = 500 feet
Width = (1000 - 2x)/2 = (1000 - 1000)/2 = 0 feet
The maximum enclosed area is:
Maximum Area = 500 × 0 = 0 square feet
However, it's important to note that since the width is zero, it means there is no enclosed area. This implies that the maximum area cannot be achieved with the given constraints of the problem, i.e., the length of the side parallel to the river being x feet and a total of 1000 feet of fencing.