Find the length and width of a rectangle that has a perimeter of 80 meters and a maximum area.
length= 10 m
width=20 m
length=10 m
width= 20 m
Find the length and width of a rectangle that has a perimeter of 80 meters and a maximum area.
length= 10 m
width= 30 m
you keep saying 10x30, but have you considered a 20x20?
10*30 < 20*20
A = x * y
2 x + 2 y = 80
x + y = 40 so y = 40 - x
A =x * (40 -x)
A =40x - x^2
dA/dx = 40 - 2 x
0 when x = 20, then y also = 20
To find the length and width of the rectangle with a maximum area, given its perimeter, we can use the principles of calculus.
Let's denote the length of the rectangle as L and the width as W. The perimeter of a rectangle is given by the formula: P = 2L + 2W.
So, in this case, we have:
2L + 2W = 80
From this equation, we can express one variable in terms of the other. Let's solve for L:
2L = 80 - 2W
L = 40 - W
Now, to find the maximum area, we use the formula for the area of a rectangle: A = L * W.
Substituting our expression for L, we have:
A = (40 - W) * W
A = 40W - W²
To find the maximum area, we need to find the maximum value of A with respect to W. We can do this by taking the derivative of A with respect to W, setting it equal to zero, and solving for W.
dA/dW = 40 - 2W = 0
Solving this equation yields:
W = 20
Now, we substitute this value of W back into the equation 2L + 2W = 80 to find the value of L:
2L + 2(20) = 80
2L + 40 = 80
2L = 40
L = 20
Therefore, the length and width of the rectangle with a perimeter of 80 meters and maximum area are 20 meters and 20 meters, respectively.