analytical solution\nfind the dimensions and area of the largest rectangle that can be inscribed in the…

Explanation:

Step1: Set up the variables

Let the coordinates of the upper - right vertex of the rectangle be ((x,y)) on the ellipse (\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1), (y\geq0). The length of the rectangle (L = 2x) and the width (W=y). The area of the rectangle (A=2xy). From the ellipse equation (y = b\sqrt{1-\frac{x^{2}}{a^{2}}}), so (A(x)=2xb\sqrt{1 - \frac{x^{2}}{a^{2}}}= \frac{2b}{a}\sqrt{a^{2}x^{2}-x^{4}}), (0\leq x\leq a).

Step2: Differentiate the area function

Let (u = a^{2}x^{2}-x^{4}), then (\frac{du}{dx}=2a^{2}x - 4x^{3}). Using the chain rule, (A^\prime(x)=\frac{2b}{a}\cdot\frac{2a^{2}x - 4x^{3}}{2\sqrt{a^{2}x^{2}-x^{4}}}=\frac{b(2a^{2}x - 4x^{3})}{a\sqrt{a^{2}x^{2}-x^{4}}}). Set (A^\prime(x)=0), then (2a^{2}x - 4x^{3}=x(2a^{2}-4x^{2}) = 0). Solutions are (x = 0) (minimum area) and (x=\frac{a}{\sqrt{2}}) (critical point).

Step3: Find the value of (y)

When (x=\frac{a}{\sqrt{2}}), (y=b\sqrt{1-\frac{(\frac{a}{\sqrt{2}})^{2}}{a^{2}}}=b\sqrt{1-\frac{1}{2}}=\frac{b}{\sqrt{2}}).

Step4: Calculate the dimensions and area

The dimensions of the rectangle: length (L = 2x=\sqrt{2}a), width (W=\frac{b}{\sqrt{2}}), so the dimensions are ((\sqrt{2}a,\frac{b}{\sqrt{2}})). The area (A = 2xy), substituting (x=\frac{a}{\sqrt{2}}) and (y=\frac{b}{\sqrt{2}}), we get (A=ab).

Answer:

dimensions: (\sqrt{2}a,\frac{b}{\sqrt{2}}) area: (ab)