A man who has 144m of fencing material wishes to enclose a rectangular garden.Find the maximum area he encloses
Well, that man must really be on the fence about his garden! But don't worry, I'm here to help. To find the maximum area, we need to determine the dimensions that would give us the largest possible rectangle.
Let's break it down:
If the length of the rectangle is "L" and the width is "W," we know that the perimeter of the rectangular garden (fencing material) is given by the equation: 2L + 2W = 144m.
Let's rearrange that equation to solve for either L or W, so we can substitute it later. How about we solve for L:
2L + 2W = 144m --> 2L = 144m - 2W --> L = 72m - W
Now, we know that the area of a rectangle is given by length multiplied by width: A = L * W.
To maximize this area, we need to find the most suitable value for L and W. Since we already have L in terms of W from rearranging the equation earlier, we can substitute it into the area formula:
A = (72m - W) * W.
Now, let's find the maximum value of A by finding the vertex of the quadratic equation: y = -W^2 + 72W.
The vertex of this quadratic equation occurs at W = -b/2a, where a = -1 and b = 72. So the width, W = -72/2(-1), which simplifies to W = 36m.
Now, substitute that back into the formula for L:
L = 72m - 36m = 36m.
So, the dimensions of the rectangular garden that will enclose the maximum area using 144m of fencing material are Length = 36m and Width = 36m.
Therefore, the maximum area enclosed is A = L * W = 36m * 36m = 1296 square meters.
To find the maximum area the man can enclose with 144m of fencing material, we can use the concept of optimization.
Let's assume the length of the rectangle is "l" and the width is "w". Since the garden is rectangular, there are two equal lengths and two equal widths.
We are given that the perimeter of the garden is 144m, so we can write an equation:
2l + 2w = 144
To solve for "w," we can rearrange the equation:
2w = 144 - 2l
w = 72 - l
Now, we can express the area of the rectangle, A, in terms of "l" and "w":
A = l * w
A = l * (72 - l)
To find the maximum area, we need to optimize the equation by finding the value of "l" that maximizes the area. We can do this by taking the derivative of the area equation with respect to "l" and setting it equal to zero:
dA/dl = 72 - 2l = 0
Now, solve for "l":
72 - 2l = 0
2l = 72
l = 36
Substituting this value back into the equation for "w":
w = 72 - l
w = 72 - 36
w = 36
So, the maximum area the man can enclose with 144m of fencing material is A = 36 * 36 = 1296 square meters.
To find the maximum area of the rectangular garden, we need to determine the dimensions that will maximize the area.
Let's assume the length of the rectangular garden is L and the width is W. Since the perimeter of the garden is 144m, we can set up the equation:
2L + 2W = 144
To simplify the equation, we can divide both sides by 2:
L + W = 72
From the equation above, we can express L in terms of W:
L = 72 - W
Now we can express the area, A, of the rectangular garden in terms of the width:
A = L * W
A = (72 - W) * W
A = 72W - W^2
To find the maximum area, we can take the derivative of A with respect to W and set it equal to zero:
dA/dW = 72 - 2W = 0
Solving this equation, we find:
W = 36
Now we can substitute this value of W back into the equation for L:
L = 72 - 36
L = 36
Therefore, the maximum area of the rectangular garden occurs when the width is 36m and the length is also 36m.
To enclose the maximum area, the man should create a square garden with each side measuring 36m.
Mathematics-ii solution
144/4 = 36
36*36 = 1296 m^2
now how did I know it was square?
2x+2y = p = perimeter
so
x+y = p/2
so
y = p/2-x
A = x y = x(p/2-x)
A = (p/2)x - x^2
or
x^2 - (p/2)x = - A
x^2 -(p/2)x + (p/4)^2 = -A + p^2/16
[ x - (p/4)]^2 = - (A-p^2/16)
so vertex at area = (p/4)^2
and x = y = p/4 so square