Mathematics, 01.07.2019 08:30 ssiy

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?

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Answer from: SymiaF4980

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-

Answer from: brooke0713

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

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

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 from: Quest

The first answer is correct
Asap plz i have 7 min left on the test

Answer from: Quest

step-by-step explanation:

to find the amount of gift bags to use all the supplies, find the gcf or greatest common factor. this is the largest number which multiplies evenly into each term.

48 has factors: 1, 2, 3, 4, 6, 8, 12. 16, 24, 48

80 has factors: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80

64 has factors: 1, 2, 4, 8, 8, 16, 32, 64

the highest number factor they all share in common is 8. this means there will be 8 bags with 6 pencils, 10 markers, and 8 stickers in each bag.

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