A rectangular lot has a perimeter of 320meters determine the maximum area of the lot
width --- x
length -- y
2x+2y = 320
y = 160-x
area = xy = x(160-x)
= -x^2 + 160x
this is a downwards opening parabola with a vertex at (80, 6400)
(I assume you know how to find the vertex)
the max area is 644 m^2 , when x = 80m
that is, when the rectangle is a square, which shows that for any given perimeter, the largest rectangle is obtained when it is a square.
Well, the maximum area of the lot would be achieved when it's a square. Since a square has all sides equal, let's divide the perimeter of the lot by 4 (since there are 4 equal sides) to find the length of each side:
320 meters / 4 = 80 meters
Now, since a square has equal sides, the length and width of the rectangular lot would both be 80 meters. To find the maximum area, we can simply calculate:
80 meters * 80 meters = 6400 square meters
So, the maximum area of the lot would be 6400 square meters, which is achieved when it's a square. I hope this helps!
To determine the maximum area of a rectangular lot with a given perimeter, we can use the concept of optimization. In this case, we need to find the dimensions that will result in the largest possible area.
Let's assume the length of the rectangular lot is L and the width is W.
Step 1: Write the equation for the perimeter:
The perimeter of a rectangle is given by the formula: P = 2L + 2W
In this case, the perimeter is given as 320 meters:
320 = 2L + 2W
Step 2: Solve for one of the variables in terms of the other:
Rearrange the equation to solve for one variable in terms of the other. Let's solve for L:
2L = 320 - 2W
L = (320 - 2W)/2
L = 160 - W
Step 3: Substitute the expression for L into the formula for area:
The area of a rectangle is given by the formula: A = L * W
Substitute the expression for L into the formula:
A = (160 - W) * W
A = 160W - W^2
Step 4: Find the maximum area:
To find the maximum area, we can take the derivative of the area function with respect to the variable W, set it equal to zero, and solve for W.
dA/dW = 160 - 2W
Setting the derivative equal to zero:
160 - 2W = 0
2W = 160
W = 160/2
W = 80
So, when W = 80, the area is maximized.
Step 5: Calculate the corresponding length and maximum area:
Substitute the value of W back into the expression for L and calculate the maximum area.
L = 160 - W
L = 160 - 80
L = 80
The length is also 80.
Now, substitute the values of L and W into the area formula:
A = L * W
A = 80 * 80
A = 6400 square meters
Therefore, the maximum area of the rectangular lot is 6400 square meters when the length and width are both 80 meters.
To determine the maximum area of a rectangular lot with a given perimeter, you can use the concept of optimization. In this case, we want to find the dimensions (length and width) that will result in the maximum area.
Let's assume the length of the rectangular lot is 'l' and the width is 'w'.
The perimeter of a rectangle is given by the formula:
Perimeter = 2l + 2w
Given that the perimeter is 320 meters, we can write the equation as:
320 = 2l + 2w
Now, we can rearrange this equation to express one variable in terms of the other. Let's solve it for 'l':
2l = 320 - 2w
l = 160 - w
Next, we need to express the area in terms of a single variable. The area of a rectangle is given by the formula:
Area = length × width
Substituting the value of 'l' in terms of 'w' into the area formula:
Area = (160 - w) × w
Now, we can simplify the equation and express it as a quadratic function of 'w':
Area = 160w - w^2
To find the maximum area, we can differentiate the area function with respect to 'w' and set it equal to zero. This will help us determine the critical points:
d(Area)/dw = 160 - 2w
Setting this to zero:
160 - 2w = 0
2w = 160
w = 80
Now that we have the value of 'w', we can substitute it back into the equation for 'l' to find its value:
l = 160 - w
l = 160 - 80
l = 80
Therefore, the dimensions that will result in the maximum area are:
Length = 80 meters
Width = 80 meters
Substituting these values into the area formula:
Area = Length × Width
Area = 80 × 80
Area = 6400 square meters
Hence, the maximum area of the rectangular lot is 6400 square meters.