Question

Use the method of Lagrange multipliers to find the rectangle of largest area, with sides parallel to the axes that can be inscribed in the ellipse${\mathbf{\left(}\frac{\mathbf{x}}{\mathbf{a}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{+}{\mathbf{\left(}\frac{\mathbf{y}}{\mathbf{b}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{=}\mathbf{1}$. What is the maximum area?

Short answer

The maximum area of an inscribed triangle can have is$=2ab$

Step-by-step solution

Define area of triangle

The area of a triangle is defined as the total space occupied by a triangle's three sides in a two-dimensional plane. The basic formula for calculating the area of a triangle is half the product of its base and height.

Calculating the area of inscribed rectangle

We want area of the inscribed rectangle to be maximum. If its sides are equal to $\mathbf{2}\mathit{x}$ and $2y$, its area is: $a(x,y)=4xy$.

$$\begin{gathered} G(x,y,\lambda )=4xy+\lambda \left[{\left(\frac{x}{a}\right)}^2+{\left(\frac{y}{b}\right)}^2-1\right] \\ \frac{\partial G}{\partial x}=0 \\ =4y+\frac{2\lambda x}{a^2} \\ y=-\frac{\lambda x}{2a^2} \\ \frac{\partial G}{\partial y}=0 \\ =4x+\frac{2\lambda y}{b^2} \\ \Rightarrow 4x=-\frac{\lambda y}{2b^2} \\ =-\frac{{\lambda }^{2}x}{a^2{b}^2} \end{gathered}$$

${\mathit{\lambda }}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{}or\mathbf{}{\mathit{\lambda }}_{\mathbf{2}}\mathbf{=}\mathbf{\pm }\mathbf{2}\mathit{a}\mathit{b}$

Calculating the maximum area

Because side of a rectangular must be positive

$$\begin{gathered} y=-\frac{\lambda x}{2a^2} \\ =\pm \frac{bx}{a} \\ =\frac{bx}{a} \\ \frac{\partial G}{\partial \lambda }=0 \\ ={\left(\frac{x}{a}\right)}^2+{\left(\frac{y}{b}\right)}^2-1 \\ \frac{x^2}{a^2}+\frac{b^2{x}^2}{a^2{b}^2}=1 \\ {x}^2=\frac{a^2}{2} \\ x=\frac{a}{\sqrt{2}},y=\frac{a}{\sqrt{2}} \end{gathered}$$

So maximum area of an inscribed triangle can have is $$\begin{gathered} A=4xy \\ =2ab \end{gathered}$$