Let x, y be the length and breadth of a rectangle and 16 is the length of diagonal as shown below:

Using the Pythagorean Theorem in the right triangle formed by sides with the diagonal, we get: x² + y² = 256²

=> \(y=\sqrt{256-x^2}\) and \(A\left(x\right)=x\sqrt{256-x^2}\)

that is the function we have to maximize. Notice that x can vary between x = 0 (the rectangle collapses to a vertical line) and x = 16 (the rectangle collapses to a horizontal line), and in both extremes the area is 0. Let’s find critical points for A:

\(0=A'\left(x\right)=\sqrt{256-x^2}-x-\dfrac{x}{\sqrt{256-x^2}}\\ \Rightarrow\dfrac{x^2}{\sqrt{256-x^2}}=\sqrt{16-x^2}\\ \Rightarrow x^2=256-x^2\)

and so \(x=\sqrt{128}\) is the only critical value

Moreover, A(\(\sqrt{8}\)) = 8 > 8 so it is the maximum

Let x, y be the length and breadth of a rectangle and 16 is the length of diagonal as shown below:

Using the Pythagorean Theorem in the right triangle formed by sides with the diagonal, we get: x2 + y2 = 2562

=> y=256−x2−−−−−−−√

 and A(x)=x256−x2−−−−−−−√

that is the function we have to maximize. Notice that x can vary between x = 0 (the rectangle collapses to a vertical line) and x = 16 (the rectangle collapses to a horizontal line), and in both extremes the area is 0. Let’s find critical points for A:

0=A′(x)=256−x2−−−−−−−√−x−x256−x2−−−−−−−√⇒x2256−x2−−−−−−−√=16−x2−−−−−−√⇒x2=256−x2

and so x=128−−−√

 is the only critical value

Moreover, A(8–√

) = 8 > 8 so it is the maximum

x y 16