## Calculus: Early Transcendentals 8th Edition

The area of the largest triangle is $~~2ab$
Let $(x,y)$ be the point where the corner of the rectangle intersects the ellipse in the first quadrant. We can find an expression for $y$: $\frac{y^2}{b^2} = 1 - \frac{x^2}{a^2}$ $y^2 = \frac{b^2}{a^2}~(a^2 -x^2)$ $y = \frac{b}{a}~\sqrt{a^2 -x^2}$ We can write an expression for the area of the rectangle: $A = (2x)(2y)$ $A = 4xy$ $A = (4x)(\frac{b}{a}~\sqrt{a^2 -x^2})$ We can find $\frac{dA}{dx}$: $\frac{dA}{dx} = \frac{4b}{a}~\sqrt{a^2 -x^2}+(\frac{4xb}{a})(\frac{-x}{\sqrt{a^2-x^2}})$ $\frac{dA}{dx} = \frac{(4b)(a^2-x^2)}{a~\sqrt{a^2-x^2}} -\frac{4x^2b}{a~\sqrt{a^2-x^2}}$ $\frac{dA}{dx} = \frac{4a^2b-4x^2b}{a~\sqrt{a^2-x^2}} -\frac{4x^2b}{a~\sqrt{a^2-x^2}}$ $\frac{dA}{dx} = \frac{4a^2b-8x^2b}{a~\sqrt{a^2-x^2}} = 0$ $4a^2b-8x^2b = 0$ $8x^2b = 4a^2b$ $x^2 = \frac{a^2}{2}$ $x = \frac{a}{\sqrt{2}}$ We can find $y$: $y = \frac{b}{a}~\sqrt{a^2 -x^2}$ $y = \frac{b}{a}~\sqrt{a^2 -(\frac{a}{\sqrt{2}})^2}$ $y = \frac{b}{a}~\sqrt{\frac{a^2}{2}}$ $y = \frac{b}{\sqrt{2}}$ We can find the area of the rectangle: $A = 4xy$ $A = 4(\frac{a}{\sqrt{2}})(\frac{b}{\sqrt{2}})$ $A = 2ab$ The area of the largest triangle is $~~2ab$