Question

Finding Area Show that among all rectangles with an 8-m perimeter, the one with largest area is a square.

Short answer

For a rectangle with an 8-meter perimeter, the one with the maximum area is a square of sides 2m by 2m.

Step-by-step solution

Formulate the problem mathematically

Let's assume the rectangle's lengths are \(x\) and \(y\). Since the perimeter is given as 8m, we can write the sum of all sides as \(2x + 2y = 8\). We can simplify this equation to \(x + y = 4\). Now, the area of the rectangle is given by \(A = x*y\). The goal is to maximize the area A.

Substitute and derive the equation

Substitute \(y\) from the simplified equation with \(y = 4 - x\) and put into the area equation, getting \(A = x*(4-x)\) or \(A = 4x - x^2\). To find maximum or minimum values, we can take the derivative and set it equals to 0, resulting in \(A' = 4 - 2x = 0\).

Solve the equation for maximum value

Solving the equation gives us \(x = 2\). Subsitute back into \(x + y = 4\), we get \(y = 2\). Thus, the rectangle with maximum area is a 2m by 2m square.

Confirm the maximum area

We can verify that this solution provides a maximum by taking the second derivative of \(A\), \(A'' = -2\), which is negative indicating a maximum. Hence, it is proven that the rectangle with an 8m perimeter that has the largest area is a square.