## Calculus: Early Transcendentals 8th Edition

The rectangle of largest area that can be inscribed in a circle of radius $r$ is a square with a side of length $\sqrt{2}~r$
Let $x$ be the width of the rectangle and let $y$ be the length of the rectangle. Note that $x \gt 0$ and $y \gt 0$ A diagonal across the rectangle has the length $2r$. Then: $x^2+y^2 = (2r)^2$ $y = \sqrt{4r^2-x^2}$ We can write an expression for the area $A$: $A = xy = x\sqrt{4r^2-x^2}$ We can find the point where $A'(x) = 0$: $A(x) = x\sqrt{4r^2-x^2}$ $A'(x) = \sqrt{4r^2-x^2}-\frac{x^2}{\sqrt{4r^2-x^2}}= 0$ $\frac{4r^2-2x^2}{\sqrt{4r^2-x^2}}= 0$ $4r^2-2x^2 = 0$ $2x^2 = 4r^2$ $x^2 = 2r^2$ $x = \sqrt{2}~r$ When $0 \lt x \lt \sqrt{2}~r$, then $A'(x) \gt 0$ When $x \gt \sqrt{2}~r$, then $A'(x) \lt 0$ Thus, $x=\sqrt{2}~r$ is the point where the area $A$ is a maximum. We can find $y$: $y = \sqrt{4r^2-x^2}$ $y = \sqrt{4r^2-(\sqrt{2}~r)^2}$ $y = \sqrt{4r^2-2r^2}$ $y = \sqrt{2}~r$ The rectangle of largest area that can be inscribed in a circle of radius $r$ is a square with a side of length $\sqrt{2}~r$