Question

A rectangle is inscribed with its base on the x-axis and its upper cornerson the parabola y = : 1 - x^2. What are the dimensions of such a rectangle with thegreatest possible area?

Answer 1

One vertex of rectangle lies on the parabola. So relation between x and y-coordinate of vertex of rectangle lying on parabola is,

`y=1-x^2`

So point,

`P(x,1-x^2)`

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,

`1-x^2`

The area of rectangle is,

`A=2x\cdot(1-x^2)^{}`

Differentiate the area equation with respect to x.

`\begin{gathered} (dA)/(dx)=(d)/(dx)\lbrack2x(1-x^2)\rbrack \\ =(d)/(dx)\lbrack2x-2x^3\rbrack \\ =2-6x^2 \end{gathered}`

For maximum area, dA/dx = 0.

Determine the value of x for maximum area.

`\begin{gathered} 2-6x^2=0 \\ 6x^2=2 \\ x=\frac{1}{\sqrt[]{3}} \end{gathered}`

The negative value is neglected as dimension can never less than 0.

The width of rectangle is 2x. So,

`\begin{gathered} \text{width = 2}\cdot\frac{1}{\sqrt[]{3}} \\ =\frac{2}{\sqrt[]{3}}*\frac{\sqrt[]{3}}{\sqrt[]{3}} \\ =\frac{2\sqrt[]{3}}{3} \end{gathered}`

The height of rectangle is 1 - x^2. So,

`\begin{gathered} \text{Height = 1-(}\frac{1}{\sqrt[]{3}})^2 \\ =1-(1)/(3) \\ =(2)/(3) \end{gathered}`

Answer:

Width =

`\frac{2\sqrt[]{3}}{3}`

Height:

`(2)/(3)`