A rectangle has an area of A. Find the dimensions that minimize the perimeter. Show that it really is a minimum.
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To find the dimensions of the rectangle that minimize the perimeter, we need to set up an equation for the perimeter in terms of one variable. Let's assume the length of the rectangle is L and the width is W.
The perimeter of a rectangle is given by the formula:
Perimeter = 2L + 2W
We also know that the area of the rectangle is given by the formula:
Area = L * W = A
To eliminate one variable, we can solve the area equation for L in terms of W:
L = A / W
Now, substitute this expression for L into the perimeter equation:
Perimeter = 2(A / W) + 2W
Simplifying further, we get:
Perimeter = 2A / W + 2W
To find the dimensions that minimize the perimeter, we can take the derivative of the perimeter equation with respect to W and set it equal to zero. This will give us the critical points where the minimum occurs.
Let's differentiate the expression for Perimeter with respect to W:
d(Perimeter) / dW = -2A / W^2 + 2
Setting this derivative equal to zero and solving for W:
-2A / W^2 + 2 = 0
-2A / W^2 = -2
A / W^2 = 1
A = W^2
Now, substitute this value of A back into the area equation:
W^2 = A
W = √A
Substitute this value of W back into the expression for L:
L = A / W = A / √A = √A
So, the dimensions that minimize the perimeter are L = √A and W = √A.
To show that this is indeed a minimum, we need to take the second derivative of the perimeter equation with respect to W and check its sign at this critical point.
Taking the second derivative:
d²(Perimeter) / dW² = 4A / W^3
At W = √A, this becomes:
d²(Perimeter) / dW² = 4A / (√A)^3 = 4A / A^(3/2) = 4 / √A
Since A is a positive value, the second derivative is always positive, indicating a minimum. Therefore, the dimensions L = √A and W = √A indeed minimize the perimeter of the rectangle.