You have 100yds. of fencing to enclose a rectangular area. Find the dimensions which maximize the area. What is the maximum area?
I know that A=L times W.
I know that 100 = 2L+2w
Subtract 2L from both sides:
100-2L=2W
Divide by two: 50-L=W
Substitute in original equation:
A=LW A=L(50-L)
A= 50L - L squared Now I am stuck.
Considering all rectangles with the same perimeter, the square encloses the greatest area.
Proof: Consider a square of dimensions x by x, the area of which is x^2. Adjusting the dimensions by adding a to one side and subtracting a from the other side results in an area of (x + a)(x - a) = x^2 - a^2. Thus, however small the dimension "a" is, the area of the modified rectangle is always less than the square of area x^2.
To find the dimensions that maximize the area, you need to maximize the equation A = L(50 - L).
To do this, you can take the derivative of this equation with respect to L and set it equal to zero to find the critical points.
dA/dL = 50 - 2L
Setting this equal to zero:
50 - 2L = 0
Solving for L:
-2L = -50
L = 25
So the critical point is L = 25.
To determine if this is a maximum or minimum, you can take the second derivative:
d²A/dL² = -2
Since the second derivative is negative, this confirms that L = 25 is a maximum.
To find the corresponding width, you can use the equation W = 50 - L:
W = 50 - 25
W = 25
Therefore, the dimensions that maximize the area are L = 25 and W = 25.
To find the maximum area, substitute these values back into the area equation:
A = L * W
A = 25 * 25
A = 625 square yards
So, the maximum area is 625 square yards.