One vertex of rectangle lies on the parabola. So relation between x and y-coordinate of vertex of rectangle lying on parabola is,
[tex]y=1-x^2[/tex]So point,
[tex]P(x,1-x^2)[/tex]lies on the rectangle.
The base of rectangle lies on the x-axis and parabola is symetric about y-axis. So width of reactangle is 2x and length of rectangle is,
[tex]1-x^2[/tex]The area of rectangle is,
[tex]A=2x\cdot(1-x^2)^{}[/tex]Differentiate the area equation with respect to x.
[tex]\begin{gathered} \frac{dA}{dx}=\frac{d}{dx}\lbrack2x(1-x^2)\rbrack \\ =\frac{d}{dx}\lbrack2x-2x^3\rbrack \\ =2-6x^2 \end{gathered}[/tex]For maximum area, dA/dx = 0.
Determine the value of x for maximum area.
[tex]\begin{gathered} 2-6x^2=0 \\ 6x^2=2 \\ x=\frac{1}{\sqrt[]{3}} \end{gathered}[/tex]The negative value is neglected as dimension can never less than 0.
The width of rectangle is 2x. So,
[tex]\begin{gathered} \text{width = 2}\cdot\frac{1}{\sqrt[]{3}} \\ =\frac{2}{\sqrt[]{3}}\times\frac{\sqrt[]{3}}{\sqrt[]{3}} \\ =\frac{2\sqrt[]{3}}{3} \end{gathered}[/tex]The height of rectangle is 1 - x^2. So,
[tex]\begin{gathered} \text{Height = 1-(}\frac{1}{\sqrt[]{3}})^2 \\ =1-\frac{1}{3} \\ =\frac{2}{3} \end{gathered}[/tex]Answer:
Width =
[tex]\frac{2\sqrt[]{3}}{3}[/tex]Height:
[tex]\frac{2}{3}[/tex]