Question

Of all rectangles with a perimeter of 12, which one has the maximum area?

Answer 1

To find the rectangle with the maximum area out of all rectangles with a perimeter of 12, we can use calculus. By taking the derivative of the area function and setting it equal to 0, we can find the critical points. After solving for the dimensions of the rectangle with the maximum area, we find that it has dimensions 3 by 3.

Step-by-step explanation:

To find the rectangle with the maximum area, we need to consider all rectangles with a perimeter of 12. Let's start by writing an equation to represent the perimeter of a rectangle: 2l + 2w = 12, where l is the length and w is the width. We can solve this equation for l in terms of w: l = 6 - w. The area of a rectangle is given by the product of its length and width, so we can substitute the value of l in terms of w into the area formula: A = lw = (6-w)w = 6w - w^2. To find the maximum area, we can take the derivative of the area function with respect to w and set it equal to 0, then solve for w. Taking the derivative, we get dA/dw = 6 - 2w = 0. Solving this equation, we find w = 3. Substituting this value back into the equation for l, we get l = 6 - 3 = 3. Therefore, the rectangle with dimensions 3 by 3 has the maximum area of 9 square units.

Answer 2

The rectangle with the maximum area has a length of 3 and a width of 3, forming a square.

Step-by-step explanation:

Let the length and width of the rectangle be "l" and "w", respectively. We know the perimeter is 12, so:

2l + 2w = 12

Solving for one variable (w) in terms of the other (l):

w = 6 - l

Now, we want to maximize the area (A) of the rectangle:

A = l * w

Substituting the expression for w:

A = l * (6 - l)

Expanding the expression:

A = 6l - l^2

To find the maximum area, we need to find the vertex of this quadratic function. We can do this by taking the derivative of A with respect to l and setting it equal to zero:

dA/dl = 6 - 2l = 0

Solving for l:

l = 3

Substituting this value back into the expression for w:

w = 6 - l = 3

Therefore, the rectangle with the maximum area has a length and width of 3, making it a square.