Find the value of x.

Mathematics

1 answer:

inessss [21]3 years ago

5 0

Answer:

Step-by-step explanation:

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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 dimension

insens350 [35]

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

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