Question

find the shortest and longest distance from the point (1,2,-1) to the sphere x^2+y^2+z^2=24 using lagrange's method of constrained maxima and minima.

Answer 1

Answer:

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Answer 2

Answer:

The shortest and longest distance from the point $(1,2,-1)$ to the sphere $x^{2} +y^{2}+z^{2}=24$ using Lagrange's method of constrained maxima and minima is $\sqrt{6}$ and $3\sqrt{6}$ respectively.

Step-by-step explanation:

The question is to find the shortest and longest distance from the point $(1,2,-1)$ to the sphere $x^{2} +y^{2}+z^{2}=24$ using Lagrange's method of constrained maxima and minima.

Consider A is the given point and P is the point on the sphere. That is,

$A(1,2,-1)$ and $P=x^{2} +y^{2} +z^{2}=24$    →(1)

The constrained function is given as follows,

$\phi \left( x,y,z\right) =x^{2}+y^{2}+z^{2}-24=0$     →(2)

The formula to find the distance between two points is,

$AP=\sqrt{\left( x_{2}-x_{1}\right) ^{2}+\left( y_{2}-y_{1}\right) ^{2}+\left( z_{2}-z_{1}\right) ^{2}}$

So the by the distance formula,

$AP=\sqrt{\left( x-1\right) ^{2}+\left( y-2\right) ^{2}+\left( z+1\right) ^{2}}$

$AP^{2}=\left( x-1\right) ^{2}+\left( y-2\right) ^{2}+\left( z+1\right) ^{2}$

So now the objective function is,  

$f\left( x,y,z\right) =\left( x-1\right) ^{2}+\left( y-2\right) ^{2}+\left( z+1\right) ^{2}$   →(3)

Now by Lagrange's method of multiplication,

$f=f\left( x,y,z\right) +\lambda \phi \left( x,y,z\right)$

$f=\left( x-1\right) ^{2}+\left( y-2\right) ^{2}+\left( z+1\right) ^{2}+\lambda \left( x^{2}+y^{2}+z^{2}-24\right)$    →(4)

We are going to partially differentiate this equation (4) with respect to x,y, and z. Then we get as follows,

$\begin{array}{l}\frac{\partial f}{\partial x}=2\left(x-1\right)+2\lambda x\\\\\frac{\partial f}{\partial y}=2\left(y-2\right)+2\lambda y\\\\\frac{\partial f}{\partial z}=2\left(z+1\right)+2\lambda z\end{array}$

Now put,  $\dfrac{\partial f}{\partial x}=0;\dfrac{\partial f}{\partial y}=0;\dfrac{\partial f}{\partial z}=0$  then,

             $\begin{array}{l}x-1+\lambda x=0\rightarrow\left(5\right)\\\\y-2+\lambda y=0\rightarrow\left(6\right)\\\\z+1+\lambda z=0\rightarrow \left( 7\right) \end{array}$

Now from equations 5,6,7, we get,

$\lambda =\dfrac{-\left( x-1\right) }{x}=\dfrac{-\left( y-2\right) }{y}=\dfrac{-\left( z+1\right) }{z}$

These three are in the same ratio. So we can write this as,

$\begin{array}{l}\lambda=\pm\frac{\sqrt{\left(x-1\right)^2+\left(y-2\right)^2+\left(z+1\right)^2}}{\sqrt{x^2+y^2+z^2}}\end{array}$

From equations (3) and (1),

$\lambda=\pm\frac{\sqrt{f}}{\sqrt{24}}$

Now put the value of $\begin{array}{l}\lambda\end{array}$ in equations 5,6, and 7, then we get,

$\begin{array}{l}x-1\pm\frac{\sqrt{f}}{\sqrt{24}}=0\\\\y-2\pm\frac{\sqrt{f}}{\sqrt{24}}=0\\\\z+1\pm\frac{\sqrt{f}}{\sqrt{24}}=0\end{array}$

Rearrange the equations to find the values of x,y, and z,

$\begin{array}{l}x=\frac{\sqrt{24}}{\sqrt{24}\pm\sqrt{f}}\\\\y=\frac{2\sqrt{24}}{\sqrt{24}\pm\sqrt{f}}\\\\z=\frac{-\sqrt{24}}{\sqrt{24}\pm \sqrt{f}}\end{array}$

From equation (1) we can write this as,

$\left( \dfrac{\sqrt{24}}{\sqrt{24}\pm \sqrt{f}}\right) ^{2}+\left(\dfrac{2\sqrt{24}}{\sqrt{24}\pm \sqrt{f}}\right) ^{2}+\left( \dfrac{-\sqrt{24}}{\sqrt{24}\pm \sqrt{f}}\right) ^{2}=24$

$\dfrac{24+4\times24+24}{\left(\sqrt{24}\pm\sqrt{f}\right)^2}=24$

$\begin{array}{l}\dfrac{24+96+24}{24}=\left(\sqrt{24}\pm\sqrt{f}\right)^2\\\\\left( \sqrt{24}\pm \sqrt{f}\right) ^{2}=6\end{array}$

$\sqrt{24}\pm \sqrt{f}=\pm \sqrt{6}$  →(8)

So in this equation (8), the + and - are on both sides. So we have to find all the possible values. That is,

$\begin{array}{l}\sqrt{f}=\sqrt{6}+\sqrt{24}=3\sqrt{6} \\\\\sqrt{f}=\sqrt{6}-\sqrt{24}=-\sqrt{6} \\\\\sqrt{f}=-\sqrt{6}+\sqrt{24}=\sqrt{6} \\\\\sqrt{f}=-\sqrt{6}-\sqrt{24}=-3\sqrt{6} \end{array}$

From the distance and objection formula, we know that $AP =\sqrt{f}$.

Since negative values can not be the distance. So the longest distance is,

$\begin{array}{l}\sqrt{f}=3\sqrt{6}\\\\\Rightarrow AP=3\sqrt{6}\end{array}$

And the shortest distance is,

$\begin{array}{l}\sqrt{f}=\sqrt{6}\\\\\Rightarrow AP=\sqrt{6}\end{array}$

This is the answer for the question.