Mathematics: Arectangle 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?

Arectangle is constructed with its base on the

Ответ:

alyssa9778

11.03.2022

Mathematics

The dimensions of the rectangle with maximum area is given by x and y

values at the maximum value of the function for its area.

The correct responses are;

$\mathrm{The \ width \ of \ the \ rectangle \ is} \ \underline{\dfrac{22}{3} \cdot \sqrt{3}}$ The height of the rectangle is $80.\overline 6$

Reasons:

The given function on which the vertices is located is y = 121 - x²

The coefficient of x in the given function is 0, therefore, the location of the

x-coordinate of the vertex of the parabola is on the y-axis.

The height of the rectangle, y = 121 - x²

The rectangle will extend equally on both sides of the y-axis, therefore;

The width of the rectangle = 2·x

The area of the rectangle, A = 2·x × (121 - x²) = 242·x - 2·x³

At the maximum area, we have;

$\dfrac{dA}{dx} =\dfrac{d \left(242 \cdot x - 2 \cdot x^3\right)}{dx} = 242 - 6 \cdot x^2 = 0$

Which gives;

$x^2 = \dfrac{242}{6}$

$x = \dfrac{11}{3} \cdot \sqrt{3}$

$\mathrm{The \ width \ of \ the \ rectangle } = 2\cdot x = 2 \times \dfrac{11}{3} \cdot \sqrt{3} = \dfrac{22}{3} \cdot \sqrt{3}$

The height of the rectangle, y = 121 - x²

$\therefore y = 121 - \dfrac{242}{6} = \dfrac{242}{3} = 80\dfrac{2}{3} = 80.\overline 6$

The height of the rectangle, y = $\underline{80.\overline 6}$

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Arectangle is constructed with its base on the x-axis and two of its vertices on the parabola y=121-

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Ответ:

Shueki

11.03.2022

Mathematics

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 $A(x_0,0)\ (x_00)$ 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

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