Question

a rectangle is constructed with its base on the x-axis and two of its vertices on the parabola y=121-x^2. what are the dimensions of the rectangle with the maximum area?

Answer 1

Answer:

the length is $\dfrac{22}{\sqrt{3}}$ and the width is $121-\left(\dfrac{11}{\sqrt{3}}\right)^2=121-\dfrac{121}{3}=\dfrac{242}{3}.$

Step-by-step explanation:

Let points A and B be placed on the x-axis. Their coordinates are $A(x_0,0)\ (x_0>0)$ and $B(-x_0,0)$ (because of parabola symmetry). Two other vertices lie on the parabola, then $C(-x_0,121-x_0^2)$ and $D(x_0,121-x_0^2).$ The length of the side AB is $2x_0$ and the length of the side AD is $121-x_0^2.$ Thus, the area of the rectangle ABCD is

$A=2x_0\cdot (121-x_0^2)=242x_0-2x_0^3.$

Find the derivative A':

$A'=242-2\cdot 3x_0^2=242-6x_0^2.$

Equate A' to 0:

$242-6x_0^2=0,\\ \\x_0^2=\dfrac{121}{3},\\ \\x_0=\dfrac{11}{\sqrt{3}}.$

The maximum area of the rectangle is

$A_{max}=242\cdot \dfrac{11}{\sqrt{3}}-2\cdot \left(\dfrac{11}{\sqrt{3}}\right)^3=\dfrac{2662}{\sqrt{3}}-\dfrac{2662}{3\sqrt{3}}=\dfrac{5324}{3\sqrt{3}}\ un^2.$

The dimensions of the rectangle are:

the length is $\dfrac{22}{\sqrt{3}}\ un.$ and the width is $121-\left(\dfrac{11}{\sqrt{3}}\right)^2=121-\dfrac{121}{3}=\dfrac{242}{3}\ un.$

Answer 2

Answer:

Equation of parabola is : y = 121 - $x^2$

Let Rectangle whose base lie on X axis have coordinates be A(a,0) and B(-a,0).

Two of vertices of rectangle lies on parabola.

If you will draw perpendicular from A(a,0) and B(-a,0) i.e on it's opposite side it's coordinate will be (a, -121 +$a^2$) and  (-a, $a^2$ - 121)

As, $(a, -121 +a^2) and (-a, a^2 - 121)$ lies on parabola : y = 121 - $x^2$.

Length = 2 a, Breadth = 121 - a²

So,A= Area of Rectangle = Length × Breadth

                              =  2 a  × (121- a²)

→A = 2 [ -a³ + 121 a]

For maxima or minima , we need to differentiate area.

A' = 2 [-3 a² + 121 ], where A = dA / da

A'= 0

2 [ - 3 a² + 121] = 0

- 3 a² + 121=0

3 a² = 121

a² = $\frac{121}{3}$

a = $\frac{11}{\sqrt3}=\frac{11\sqrt3}{3}$

Length = 2 a = $\frac{22\sqrt3}{3}\text{ and Breadth} = 121 - [\frac{11\sqrt3}{3}]^2=\frac{726}{9}$

A" = 2 [ -6 a ], where A" = d A²/ d a²

A" = - 12 a, which is negative,  gives Maximum area.