Find the dimensions of a rectangle with perimeter 100m. whose area is as large as possible.
x + y = 50
so
y = 50 -x
A = x y = x(50 - x ) = 50 x - x^2
dA/dx = 0 for max or min = 50 -2x
2 x = 50
x = 25
so it is a square 25 on a side.
let x be width and y be length ,and as we know perimeter of rectangle=2(length+width)
so 2(x+y)=100
x+y=50 , x=50-y
area=length*width
area=xy=(50-y)y
area=50y-y^2
dA/dy=0 for min or max
dA/dy=50-2y=0
y=25
x=25
let x be width and y be length ,and as we know perimeter of rectangle=2(length+widith)
so 2(x+y)=100
x+y=50 , x=50-y
area=length*width
area=xy=(50-y)y
area=50y-y^2
dA/dy=0 for min or max
dA/dy=50-2x=0
y=25
x=25
thanks a lot!
Thank you so much
Well, if we want to find the dimensions of a rectangle with a perimeter of 100 meters, we need to remember that the formula for perimeter is P = 2l + 2w, where l is the length and w is the width of the rectangle.
Now, since we want to maximize the area, we need to find the dimensions that will give us the largest possible area. The formula for area is A = l * w.
So, let's find the dimensions by solving the perimeter equation for w: w = (P - 2l) / 2.
Substituting the given perimeter of 100 meters into the equation, we get: w = (100 - 2l) / 2.
To maximize the area, we want to find the value of l that will make w as large as possible. If we set w to its maximum value, the rectangle will be a square, so w = l.
Now, substituting w = l into the equation, we get: l = (100 - 2l) / 2.
Simplifying further, we have: l = 50 - l.
So, by solving this equation, we find that l = 25.
Therefore, the dimensions of the rectangle with a perimeter of 100 meters and the largest possible area are 25m x 25m.
But remember, this is serious clown business, so the final answer is a rectangle with dimensions 25m by 25m. Just don't try to paint a smiley face on it!
To find the dimensions of a rectangle with the largest possible area given a fixed perimeter, you can use the concept of calculus and the properties of rectangles.
First, let's define the rectangle's width as "w" and its length as "l". The perimeter of a rectangle is given by the formula:
P = 2w + 2l
In this case, we are given that the perimeter is 100m. So, the equation becomes:
100 = 2w + 2l
Next, let's express the area of the rectangle in terms of w and l. The area of a rectangle is given by the formula:
A = w * l
To find the dimensions that maximize the area, we need to find a relation between w and l.
Let's solve the perimeter equation for l:
100 = 2w + 2l
100 - 2w = 2l
50 - w = l
Now, substitute this value for l in the area equation:
A = w * (50 - w)
To maximize the area, we need to find the critical points of this function. Let's differentiate the area equation with respect to w:
dA/dw = 50 - 2w
Next, set the derivative equal to zero to find the critical point:
50 - 2w = 0
2w = 50
w = 25
Now that we have a critical point, we need to determine if it yields the maximum area. We can do this by checking the second derivative. Differentiate the derivative with respect to w:
d²A/dw² = -2
Since the second derivative is negative, we can conclude that w = 25 corresponds to a maximum area.
Substitute this value back into the expression for l:
l = 50 - w
l = 50 - 25
l = 25
Therefore, the dimensions of the rectangle with a perimeter of 100m and the largest possible area are width = 25m and length = 25m.