Chapter 20 Maxima and minima of functions of two and three variables-I

Watch this video that covers the concept of maxima, minima of three variables intuitively and clearly before proceeding further.

20.0.1 Question 1

Examine for maxima and minima and obtain these values for following functions

1. \(f=x^2 -xy + y^2 + 3x - 2y + 1\)

Here,

\[\begin{equation*} \begin{split} f_x &= 2x-y+3\\ f_y &= -x + 2y-2\\ f_{xx} &= 2\\ f_{xy} &= -1\\ f_{yy} &= 2 \end{split} \end{equation*}\]

For extreme values,

\[\begin{align} f_x &= 0 \nonumber \\ 2x-y+3 &= 0 \tag{20.1} \end{align}\]

\[\begin{align} f_y &= 0 \nonumber \\ -x + 2y - 2 &= 0 \tag{20.2} \end{align}\]

Solving (20.1) and (20.2), we get \(x=-\frac{4}{3}, y=\frac{1}{3}\).

For point \(\left(-\frac{4}{3}, \frac{1}{3}\right)\),

\[\begin{equation*} \begin{split} f_{xy} &= -1 \end{split} \end{equation*}\]

Thus \(f_{xx} > 0\) and \(f_{xx}f_{yy}-(f_{xy})^2=2\times2-(-1)^2=3\) which is \(>0\). Thus \(f(x,y)\) has a minimum at \((-\frac{4}{3}, \frac{1}{3})\). Its value is

\[\begin{equation*} \begin{split} &= \frac{16}{9} + \frac{4}{9} + \frac{1}{9} - 4- \frac{2}{3} + 1\\ &= \frac{21}{9}-\left(\frac{9+2}{3}\right)\\ &= \frac{21}{9}- \frac{11}{3}\\ &= -\frac{4}{3} \end{split} \end{equation*}\]

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2. \(u=16-(x+2)^2 - (y-2)^2\)

Here,

\[\begin{equation*} \begin{split} u_x &= -2(x+2)\\ u_y &= -2(y-2)\\ u_{xx} &= -2\\ u_{xy} &= 0\\ u_{yy} &= -2 \end{split} \end{equation*}\]

For extreme values,

\[\begin{align*} u_x &= 0 \\ x &= -2 \end{align*}\]

\[\begin{align*} u_y &= 0 \\ y &= 2 \end{align*}\]

The stationary point is thus \((-2,2)\). So

\[\begin{equation*} \begin{split} u_{xx}u_{yy} -(u_{xy})^2 &= (-2)\times (-2)-0\\ &= 4 > 0 \end{split} \end{equation*}\]

Thus \(u_{xx} < 0\) and \(u_{xx}u_{yy}-(u_{xy})^2>0\). Thus \(u(x,y)\) has a local maximum at \((-2,2)\). Its value is

\[\begin{equation*} \begin{split} &= 16-0-0\\ &= 16 \end{split} \end{equation*}\]

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3. \(z=8-4x + 4y -x^2 -y^2\)

Here,

\[\begin{equation*} \begin{split} z_x &= -4-2x\\ z_y &= 4-2y\\ z_{xx} &= -2\\ z_{xy} &= 0\\ z_{yy} &= -2 \end{split} \end{equation*}\]

For extreme values,

\[\begin{align*} z_x &= 0 \\ x &= -2 \end{align*}\]

\[\begin{align*} z_y &= 0 \\ y &= 2 \end{align*}\]

The stationary point is thus \((-2,2)\). So

\[\begin{equation*} \begin{split} z_{xx}z_{yy} -(z_{xy})^2 &= (-2)\times (-2)-0\\ &= 4 > 0 \end{split} \end{equation*}\]

Thus \(z_{xx} < 0\) and \(z_{xx}z_{yy}-(z_{xy})^2>0\). Thus \(z(x,y)\) has a local maximum at \((-2,2)\). Its value is

\[\begin{equation*} \begin{split} &= 8 + 8 + 8-4-4\\ &= 16 \end{split} \end{equation*}\]

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4. \(z=x^3 - x^2 - y^2 + xy\)

Here,

\[\begin{equation*} \begin{split} z_x &= 3x^2-2x+y\\ z_y &= -2y + x\\ z_{xx} &= 6x - 2\\ z_{xy} &= 1\\ z_{yy} &= -2 \end{split} \end{equation*}\]

For extreme values,

\[\begin{align} z_x &= 0 \nonumber \\ 3x^2-2x+y &= 0 \tag{20.3} \end{align}\]

\[\begin{align} z_y &= 0 \nonumber \\ x-2y &= 0 \tag{20.4} \end{align}\]

Solving (20.3) and (20.4), we get \(x=0, \frac{1}{2}\).

The function \(z\) has extreme points at (0,0) and \((1/2,1/4)\).

At \((0,0)\),

\[\begin{equation*} \begin{split} z_{xx} &= -2\\ z_{xx}z_{yy}-(z_{xy})^2 &= (-2)\times (-2) -1\\ &= 3 \end{split} \end{equation*}\]

Thus \(z_{xx} < 0\) and \(z_{xx}z_{yy}-(z_{xy})^2>0\). Thus \(z(x,y)\) has a maximum at \((0,0)\) and its value is \(0\).

At \((1/2, 1/4)\),

\[\begin{equation*} \begin{split} z_{xx} &= 1\\ z_{xx}z_{yy}-(z_{xy})^2 &= 1 \times (-2) -1 \\ &= -3 \end{split} \end{equation*}\]

Thus \(z_{xx} > 0\) and \(z_{xx}z_{yy}-(z_{xy})^2 < 0\). Thus \(z(x,y)\) has neither a maximum nor minimum at \((1/2,1/4)\) i.e. its a saddle point.

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22. \(u=2x^2 + xy + 4y^2 + xz + z^2 + 2\)

The given function \(u\) has three variables that can be represented as \(u(x,y,z)\).

Here,

\[\begin{equation*} \begin{split} u_x &= 4x + y+z\\ u_y &= x + 8y\\ u_z &= x + 2z\\ u_{xx} &= 4\\ u_{xy} &= 1\\ u_{xz} &= 1\\ u_{yx} &= 1\\ u_{yy} &= 8\\ u_{yz} &= 0\\ u_{zx} &= 1\\ u_{zy} &= 0\\ u_{zz} &= 2\\ \end{split} \end{equation*}\]

For extreme values,

\[\begin{align} u_x &= 0 \nonumber \\ 4x + y+ z &= 0 \tag{20.5} \end{align}\]

\[\begin{align} u_y &= 0 \nonumber \\ x+8y &= 0 \tag{20.6} \end{align}\]

\[\begin{align} u_z &= 0 \nonumber \\ x+2z &= 0 \tag{20.7} \end{align}\]

Solving (20.5), (20.6) and (20.7), we get \(x=0, y=0, z=0\). The stationary point is therefore \((0,0,0)\).

Here, \(u_{xx}>0\),

\[ \begin{vmatrix} u_{xx} & u_{xy} \\ u_{yx} & u_{yy} \end{vmatrix} = \begin{vmatrix} 4 & 1\\ 1 & 8 \end{vmatrix} =31 > 0 \]

and

\[ \begin{vmatrix} u_{xx} & u_{xy} & u_{xz} \\ u_{yx} & u_{yy} & u_{yz} \\ u_{zx} & u_{zy} & u_{zz} \end{vmatrix} = \begin{vmatrix} 4 & 1 & 1 \\ 1 & 8 & 0 \\ 1 & 0 & 2 \end{vmatrix} = 54 > 0 \]

Thus \(u\) has minimum value at \((0,0,0)\) and the value is

\[\begin{equation*} \begin{split} u(0,0,0) &= 2 \end{split} \end{equation*}\]

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6. \(\phi = 35 - (2x+3)^2 - (y-4)^2 - (z+1)^2\)

The given function \(\phi\) has three variables that can be represented as \(\phi(x,y,z)\).

Here,

\[\begin{equation*} \begin{split} \phi_x &= -2(2x+3)\times 2=-4(2x+3)\\ \phi_y &= -2(y-4)\\ \phi_z &= -2(z+1)\\ \phi_{xx} &= -8\\ \phi_{xy} &= 0\\ \phi_{xz} &= 0\\ \phi_{yx} &= 0\\ \phi_{yy} &= -2\\ \phi_{yz} &= 0\\ \phi_{zx} &= 0\\ \phi_{zy} &= 0\\ \phi_{zz} &= -2\\ \end{split} \end{equation*}\]

For extreme values,

\[\begin{equation*} \begin{split} \phi_x &= 0\\ -4(2x+3) &= 0\\ x &= -\frac{3}{2} \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \phi_y &= 0\\ -2(y-4) &= 0\\ y &= 4 \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \phi_z &= 0\\ -2(z+1) &= 0\\ z &= -1 \end{split} \end{equation*}\]

The stationary point is thus \(\left(-\frac{3}{2},4,-1\right)\).

Here, \(\phi_{xx}=-8 < 0\),

\[ \begin{vmatrix} \phi_{xx} & \phi_{xy} \\ \phi_{yx} & \phi_{yy} \end{vmatrix} = \begin{vmatrix} -8 & 0\\ 0 & -2 \end{vmatrix} =16 > 0 \]

and

\[ \begin{vmatrix} \phi_{xx} & \phi_{xy} & \phi_{xz} \\ \phi_{yx} & \phi_{yy} & \phi_{yz} \\ \phi_{zx} & \phi_{zy} & \phi_{zz} \end{vmatrix} = \begin{vmatrix} -8 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2 \end{vmatrix} = -32 < 0 \]

Thus \(\phi\) has maximum value at \((-3/2,4,-1)\) and the value is

\[\begin{equation*} \begin{split} \phi\left(-\frac{3}{2},4,-1\right) &= 35-0-0-0 &= 35 \end{split} \end{equation*}\]

20.0.2 Question 2

Examine for maxima and minima of following functions

1. \(2x^2 - 3xy + 4y^2 + 5\) (TU 2065)

Let \(u=2x^2 - 3xy + 4y^2 + 5\), this is a function of two variables \(u(x,y)\).

\[\begin{equation*} \begin{split} u_x &= 4x-3y\\ u_y &= -3x+8y \\ u_{xx} &= 4\\ u_{xy} &= -3\\ u_{yy} &= 8\\ u_{yx} &= -3 \end{split} \end{equation*}\]

For extreme values,

\[\begin{align} u_x &= 0\\ 4x-3y &= 0 \tag{20.8} \end{align}\]

\[\begin{align} u_y &= 0\\ -3x + 8y &= 0 \tag{20.9} \end{align}\]

Solving (20.8) and (20.9), we get \(x=0, y=0\). The stationary point is thus \((0,0)\).

Here, \(u_{xx}=4 > 0\) and

\[ \begin{vmatrix} u_{xx} & u_{xy} \\ u_{yx} & u_{yy} \end{vmatrix} = \begin{vmatrix} 4 & -3\\ -3 & 8 \end{vmatrix} = 32-9 =23 > 0 \]

So \(u\) has minimum at \((0,0)\) and the value is \(5\).

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2. \(xy + \frac{a^3}{x} + \frac{a^3}{y}\)

Let \(f=xy + \frac{a^3}{x}+\frac{a^3}{y}\) which is a function of two variables \(f(x,y)\).

\[\begin{equation*} \begin{split} f_x &= y-\frac{a^3}{x^2}\\ f_y &= x - \frac{a^3}{y^2}\\ f_{xx} &= \frac{2a^3}{x^3}\\ f_{yy} &= \frac{2a^3}{y^3} \end{split} \end{equation*}\]

For extreme values,

\[\begin{align} f_x &= 0\\ y-\frac{a^3}{x^2} &= 0\\ x^2y &= a^3 \tag{20.10} \end{align}\]

\[\begin{align} f_y &= 0\\ x - \frac{a^3}{y^2} &= 0\\ xy^2 &= a^3 \tag{20.11} \end{align}\]

Solving (20.10) and (20.11), we get \(x=a, y=a\). The stationary point is therefore \((a,a)\). At point \((a,a)\), \(f_{xx}=2 > 0\) and \(f_{yy} = 2\) and

\[ \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix} = \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} = 3 > 0 \]

Thus function has minimum at \((a,a)\) and the value is \(a^2 + a^2 + a^2=3a^2\).

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3. \(x^3 + y^3 - 12x -3y + 20\)

Let \(f=x^3 + y^3 - 12x -3y + 20\) which is a function of two variables \(f(x,y)\).

\[\begin{equation*} \begin{split} f_x &= 3x^2 - 12\\ f_y &= 3y^2 - 3\\ f_{xx} &= 6x\\ f_{yy} &= 6y\\ f_{xy} &= 0\\ f_{yx} &= 0 \end{split} \end{equation*}\]

For extreme values,

\[\begin{equation*} \begin{split} f_x &= 0\\ 3x^2 - 12 &= 0\\ x &= \pm 2\\ f_y &= 0\\ 3y^2 - 3 &= 0\\ y &= \pm 1 \end{split} \end{equation*}\]

Thus there are four possible extreme points \((2,1), (2,-1), (-2,1)\) and \((-2,-1)\).

At \((2,1)\),

\[\begin{equation*} \begin{split} f_{xx} &= 12 > 0\\ f_{yy} &= 6 \end{split} \end{equation*}\]

\[ \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix} = \begin{vmatrix} 12 & 0 \\ 0 & 6 \end{vmatrix} = 72 > 0 \]

Thus \(f\) has a minimum at \((2,1)\) and value is \(8+1-24-3+20=2\).

At \((2,-1)\),

\[\begin{equation*} \begin{split} f_{xx} &= 12 > 0\\ f_{yy} &= -6 \end{split} \end{equation*}\]

and

\[ \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix} = \begin{vmatrix} 12 & 0 \\ 0 & -6 \end{vmatrix} = -72 < 0 \]

It has neither maximum nor minimum at \((2,-1)\) i.e. it is a saddle point.

At \((-2,1)\),

\[\begin{equation*} \begin{split} f_{xx} &= -12 < 0\\ f_{yy} &= 6 \end{split} \end{equation*}\]

and

\[ \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix} = \begin{vmatrix} -12 & 0 \\ 0 & 6 \end{vmatrix} = -72 < 0 \]

So there is saddle point at \((-2,1)\).

At \((-2,-1)\),

\[\begin{equation*} \begin{split} f_{xx} &= -12 < 0\\ f_{yy} &= -6 \end{split} \end{equation*}\]

and

\[ \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix} = \begin{vmatrix} -12 & 0 \\ 0 & -6 \end{vmatrix} = 72 >0 \]

Thus there is maximum at \((-2,-1)\) and the maximum value is \(f(-2,-1)=-8-1+24+3+20=38\).

Figure 20.1: Plotting of \(x^3 + y^3 - 12x -3y + 20\) showing two saddle points, one maximum and one minimum

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4. \(y^2 + 2x^2 - 5x^4 + 4x^5\)

Let \(f=y^2 + 2x^2 - 5x^4 + 4x^5\) which is a function of two variables \((x,y)\).

\[\begin{equation*} \begin{split} f_x &= 4x-20x^3 + 20x^4\\ f_y &= 2y\\ f_{xx} &= 4-60x^2 + 80x^3\\ f_{xy} &= 0\\ f_{yx} &= 0\\ f_{yy} &= 2 \end{split} \end{equation*}\]

For extreme values ,

\[\begin{align} f_x &= 0\\ 4x-20x^3 + 20x^4 &= 0\\ x(1-5x^2+5x^3) &= 0 \tag{20.12} \end{align}\]

\[\begin{align} f_y &= 0\\ 2y &= 0\\ y &= 0 \tag{20.13} \end{align}\]

From (20.12), there are two real solutions and two imaginary solutions.

Real solutions are \(x=0, - \frac{\sqrt[3]{\frac{3 \sqrt{21}}{10} + \frac{17}{10}}}{3} - \frac{1}{3 \sqrt[3]{\frac{3 \sqrt{21}}{10} + \frac{17}{10}}} + \frac{1}{3}\)

Imaginary solutions are \(x=\frac{1}{3} - \frac{1}{3 \left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{3 \sqrt{21}}{10} + \frac{17}{10}}} - \frac{\left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{3 \sqrt{21}}{10} + \frac{17}{10}}}{3},\\ \frac{1}{3} - \frac{\left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{3 \sqrt{21}}{10} + \frac{17}{10}}}{3} - \frac{1}{3 \left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{3 \sqrt{21}}{10} + \frac{17}{10}}}\)

Solving \(1-5x^2 + 5x^3\) is cumbersome manually, so I solved it with sympy in python with the following code

from sympy.solvers import solve
from sympy import *
x = Symbol('x')

solutions=solve(1 - 5*x**2 + 5*x**3 , x, dict=True)
for i in solutions:
    print(latex(i.get(x)))

For any \(x\) value, \(y=0\).

Taking \((0,0)\),

\[\begin{equation*} \begin{split} f_{xx} &= 4 >0\\ f_{yy} &= 2\\ f_{xy} &= 0\\ f_{yx} &= 0 \end{split} \end{equation*}\]

and

\[ \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix} = \begin{vmatrix} 4 & 0 \\ 0 & 2 \end{vmatrix} =8 > 0 \]

Thus the given function with two variables \((x,y)\) have minimum at \((0,0)\).

Figure 20.2: Plotting of \(y^2+ 2x^2-5x^4+4x^5\)

Plotting shows minimum at \((0,0)\).

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22. \(-3x^2 + 6xz + 4y - 2y^2 - 6z^2\)

The given function has three variables. Let \(u(x,y,z)=-3x^2 + 6xz + 4y - 2y^2 - 6z^2\).

Here,

\[\begin{equation*} \begin{split} u_x &= -6x+6z\\ u_y &= 4-4y\\ u_z &= 6x-12z\\ u_{xx} &= -6\\ u_{xy} &= 0\\ u_{xz} &= 6\\ u_{yx} &= 0\\ u_{yy} &= -4\\ u_{yz} &= 0\\ u_{zx} &= 6\\ u_{zy} &= 0\\ u_{zz} &= -12\\ \end{split} \end{equation*}\]

For extreme values,

\[\begin{align} u_x &= 0 \nonumber \\ -6x + 6z &= 0 \nonumber \\ -x + z &= 0 \tag{20.14} \end{align}\]

\[\begin{align} u_y &= 0 \nonumber \\ 4-4y &= 0 \nonumber \\ 1-y &= 0 \tag{20.15} \end{align}\]

\[\begin{align} u_z &= 0 \nonumber \\ 6x-12z &= 0 \nonumber \\ x-2z &= 0 \tag{20.16} \end{align}\]

Solving (20.14), (20.15) and (20.16), we get \(x=0, y=1, z=0\). The stationary point is therefore \((0,1,0)\).

Here, \(u_{xx}=-6<0\),

\[ \begin{vmatrix} u_{xx} & u_{xy} \\ u_{yx} & u_{yy} \end{vmatrix} = \begin{vmatrix} -6 & 0\\ 0 & -4 \end{vmatrix} =24 > 0 \]

and

\[ \begin{vmatrix} u_{xx} & u_{xy} & u_{xz} \\ u_{yx} & u_{yy} & u_{yz} \\ u_{zx} & u_{zy} & u_{zz} \end{vmatrix} = \begin{vmatrix} -6 & 0 & 6 \\ 0 & -4 & 0 \\ 6 & 0 & -12 \end{vmatrix} = -144 < 0 \]

Thus \(u\) has maximum value at \((0,1,0)\) and the value is

\[\begin{equation*} \begin{split} u(0,1,0) &= 4-2 =2 \end{split} \end{equation*}\]

20.0.3 Question 3

Find the maximum and minimum values of

1. \(x^3y^2(1-x-y), x \neq 0, y \neq 0, x+y \neq 1\)

Here \(f=x^3y^2(1-x-y), x \neq 0, y \neq 0, x+y \neq 1\).

\[\begin{equation*} \begin{split} f_x &= y^2[x^3.(-1)+(1-x-y).3x^2]\\ &= -x^3y^2 + 3x^2y^2(1-x-y)\\ f_{yx} &= -2x^3y + 3x^2[y^2.(-1) + (1-x-y).2y]\\ &= -2x^3y - 3x^2y^2 + 6x^2y(1-x-y)\\ f_y &= x^3[y^2.(-1)+(1-x-y).2y]\\ &= -x^3y^2 + 2x^3y(1-x-y)\\ f_{xy} &= -3x^2y^2 + 2y[x^3.(-1)+(1-x-y).(3x^2)]\\ &= -3x^2y^2-2x^3y + 6x^2y(1-x-y)\\ &= -2x^3y - 3x^2y^2 + 6x^2y(1-x-y)\\ f_{xx} &= -3x^2y^2 + 3y^2[x^2.(-1)+(1-x-y).(2x)]\\ &= -6x^2y^2 + 6xy^2(1-x-y)\\ f_{yy} &= -2x^3y + 2x^3[y.(-1)+(1-x-y)]\\ &= -4x^3y + 2x^3(1-x-y) \end{split} \end{equation*}\]

For extreme values,

\[\begin{align} f_x &= 0\\ -x^3y^2 + 3x^2y^2(1-x-y) &= 0 \tag{20.17} \end{align}\]

\[\begin{align} f_y &= 0\\ -x^3y^2 + 2x^3y(1-x-y) &= 0 \tag{20.18} \end{align}\]

Solving (20.17) and (20.18),

\[\begin{equation*} \begin{split} (1-x-y)(3x^2y^2-2x^3y) &= 0\\ x^2y(1-x-y)(3y-2x) &= 0 \end{split} \end{equation*}\]

As \(x \neq 0, y \neq 0, x+y \neq 1\),

\[\begin{align} 3y -2x &= 0\\ x &= \frac{3}{2}y \tag{20.19} \end{align}\]

Substituting this in (20.17),

\[\begin{equation*} \begin{split} -\frac{27}{8}y^3y^2 +3 \times \frac{9}{4}y^2y^2\left(1-\frac{3}{2}y-y\right) &= 0\\ y^4\left(-\frac{27}{8}y+\frac{27}{4}\left(1-\frac{5y}{2}\right)\right) &= 0\\ y^4\left( y\left(\frac{-27}{8}\times 6 \right) + \frac{27}{4} \right) &= 0 \end{split} \end{equation*}\]

As \(y\neq 0\),

\[\begin{equation*} \begin{split} y\left(\frac{-27}{8}\times 6 \right) + \frac{27}{4} &= 0\\ y &= -\frac{27}{4} \times \frac{8}{-27 \times 6}\\ y &= \frac{1}{3} \end{split} \end{equation*}\]

Plugging this into (20.19),

\[\begin{equation*} \begin{split} x=\frac{1}{2} \end{split} \end{equation*}\]

The stationary point is thus \(\left(\frac{1}{2},\frac{1}{3}\right)\). At this point we have,

\[\begin{equation*} \begin{split} f_{xx} &= -6\times \frac{1}{4}\times \frac{1}{9} + 6\times \frac{1}{2}\times \frac{1}{9}\left(1-\frac{1}{2}-\frac{1}{3}\right)\\ &= -\frac{1}{6}+\frac{1}{3}\frac{6-3-2}{6}\\ &= -\frac{1}{9}\\ f_{yy} &= -4\times \frac{1}{8}\times \frac{1}{3} + 2\times \frac{1}{8}\left(1-\frac{1}{2}-\frac{1}{3}\right)\\ &= -\frac{1}{6} + \frac{1}{4}\times \frac{1}{6}\\ &= -\frac{1}{8}\\ f_{xy} &= -2\times \frac{1}{8}\times \frac{1}{3} - 3\times \frac{1}{4}\times \frac{1}{9} + 6 \times \frac{1}{4} \times \frac{1}{3}\left(1-\frac{1}{2}-\frac{1}{3}\right)\\ &= -\frac{1}{12}-\frac{1}{12}+ \frac{1}{2}\times \frac{1}{6}\\ &= -\frac{1}{12} \end{split} \end{equation*}\]

Here \(f_{xx}= -\frac{1}{9}<0\),

\[ \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix} = \begin{vmatrix} -\frac{1}{9} & -\frac{1}{12} \\ -\frac{1}{12} & -\frac{1}{8} \end{vmatrix} = \frac{1}{72}-\frac{1}{144}=\frac{1}{144} > 0 \]

So \(f\) has maximum at \(\left(\frac{1}{2},\frac{1}{3}\right)\) and the value is

\[\begin{equation*} \begin{split} f\left(\frac{1}{2},\frac{1}{3}\right) &= \frac{1}{8}\times \frac{1}{9}(1-\frac{1}{2}-\frac{1}{3})\\ &= \frac{1}{8}\times \frac{1}{9} \times \frac{1}{6}\\ &= \frac{1}{432} \end{split} \end{equation*}\]

2. \(2(x-y)^2 - x^4 - y^4\)

Let \(f=2(x-y)^2-x^4-y^4\) which is a function with two variables \(x,y\). Taking partial derivatives,

\[\begin{equation*} \begin{split} f_x &= 4(x-y)-4x^3\\ f_y &= -4(x-y)-4y^3\\ f_{xx} &= 4-12x^2\\ f_{yy} &= 4-12y^2\\ f_{xy} &= -4\\ f_{yx} &= -4 \end{split} \end{equation*}\]

For extreme values,

\[\begin{align} f_x &= 0\\ 4(x-y)-4x^3 &= 0 \tag{20.20} \end{align}\]

\[\begin{align} f_y &= 0\\ -4(x-y)-4y^3 &= 0 \tag{20.21} \end{align}\]

Solving (20.20) and (20.21), we get \(x=-y\). Plugging this into (20.20),

\[\begin{equation*} \begin{split} 4\times (-2y) + 4y^3 &= 0\\ y^3-2y &= 0\\ y(y^2-2) &= 0\\ y &= 0, +\sqrt{2}, -\sqrt{2} \end{split} \end{equation*}\]

So the points with extreme values are \((0,0), (-\sqrt{2},\sqrt{2})\) and \((\sqrt{2},-\sqrt{2})\).

At \((0,0)\),

\[\begin{equation*} \begin{split} f_{xx} &= 4 > 0\\ f_{yy} &= 4\\ f_{xy}, f_{yx} &= -4 \end{split} \end{equation*}\]

and

\[ \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix} = \begin{vmatrix} 4 & -4 \\ -4 & 4 \end{vmatrix} = 16-16 = 0 \]

The function is thus doubtful at \((0,0)\).

At \((-\sqrt{2},\sqrt{2})\),

\[\begin{equation*} \begin{split} f_{xx} &= -20 < 0\\ f_{yy} &= -20\\ f_{xy}, f_{yx} &= -4 \end{split} \end{equation*}\]

and

\[ \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix} = \begin{vmatrix} -20 & -4 \\ -4 & -20 \end{vmatrix} =384 > 0 \]

So \(f\) has maximum at \((-\sqrt{2},\sqrt{2})\).

At \((\sqrt{2},-\sqrt{2})\),

\[\begin{equation*} \begin{split} f_{xx} &= -20 < 0\\ f_{yy} &= -20\\ f_{xy}, f_{yx} &= -4 \end{split} \end{equation*}\]

and

\[ \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix} = \begin{vmatrix} -20 & -4 \\ -4 & -20 \end{vmatrix} =384 > 0 \]

So \(f\) has maximum at \((\sqrt{2},-\sqrt{2})\).

The function has thus maximum at two points \((-\sqrt{2},\sqrt{2})\) and \((\sqrt{2},-\sqrt{2})\) and the value is \(8\) at both points.

The plotting indeed proves this.

Figure 20.3: Plotting of \(2(x-y)^2-x^4-y^4\)

3. \(x^3 + 3xy^2 - 15x^2 - 15y^2 + 72x\)

This is a function of two variables \(f(x,y) = x^3 + 3xy^2 - 15x^2 - 15y^2 + 72x\).

\[\begin{equation*} \begin{split} f_x &= 3x^2+3y^2-30x+72\\ f_y &= 6xy-30y\\ f_{xx} &= 6x-30\\ f_{yy} &= 6x-30\\ f_{xy} &= 6y\\ f_{yx} &= 6y \end{split} \end{equation*}\]

For extreme values,

\[\begin{align} f_x &= 0\\ 3x^2+3y^2-30x+72 &= 0\\ x^2 +y^2 -10x + 24 &= 0 \tag{20.22} \end{align}\]

\[\begin{align} f_y &= 0\\ 6xy-30y &= 0\\ y(x-5) &= 0 \tag{20.23} \end{align}\]

Solving (20.22) and (20.23), when \(y=0\), \(x=4,6\) and when \(x=5\), \(y=\pm 1\). So the extreme points are \((4,0), (6,0),(5,1)\) and \((5,-1)\).

At \((4,0)\),

\[\begin{equation*} \begin{split} f_{xx} &= -6 <0\\ f_{yy} &= -6\\ f_{xy} &= 0\\ f_{yx} &= 0 \end{split} \end{equation*}\]

\[\begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix}\] = \[\begin{vmatrix} -6 & 0 \\ 0 & -6 \end{vmatrix}\]

=36 > 0

So there is maximum at this point and the value is \(64-240+288=112\).

At \((6,0)\),

\[\begin{equation*} \begin{split} f_{xx} &= 6 > 0\\ f_{yy} &= 6\\ f_{xy} &= 0\\ f_{yx} &= 0 \end{split} \end{equation*}\]

\[\begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix}\] = \[\begin{vmatrix} 6 & 0 \\ 0 & 6 \end{vmatrix}\]

=36 > 0

Thus there is minimum at \((6,0)\) and the value is \(216-540+432=108\).

At \((5,1)\),

\[\begin{equation*} \begin{split} f_{xx} &= 0\\ f_{yy} &= 0\\ f_{xy} &= 6\\ f_{yx} &= 6 \end{split} \end{equation*}\]

\[\begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix}\] = \[\begin{vmatrix} 0 & 6 \\ 6 & 0 \end{vmatrix}\]

=-36 < 0

At \((5,-1)\),

\[\begin{equation*} \begin{split} f_{xx} &= 0\\ f_{yy} &= 0\\ f_{xy} &= -6\\ f_{yx} &= -6 \end{split} \end{equation*}\]

\[\begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix}\] = \[\begin{vmatrix} 0 & -6 \\ -6 & 0 \end{vmatrix}\]

=-36 < 0

The function has saddle at \((5,1)\) and \((5,-1)\).

Figure 20.4: Plotting of \(x^3+3xy^2-15x^2-15y^2+72x\)

4. \(8z + 2x^2 + 3y^2 + 4z^2 - 3xy\)

The given function \(u\) has three variables that can be represented as \(u(x,y,z)\).

Here,

\[\begin{equation*} \begin{split} u_x &= 4x -3y\\ u_y &= 6y-3x\\ u_z &= 8 + 8z\\ u_{xx} &= 4\\ u_{xy} &= -3\\ u_{xz} &= 0\\ u_{yx} &= -3\\ u_{yy} &= 6\\ u_{yz} &= 0\\ u_{zx} &= 0\\ u_{zy} &= 0\\ u_{zz} &= 8\\ \end{split} \end{equation*}\]

For extreme values,

\[\begin{align} u_x &= 0 \nonumber \\ 4x -3y &= 0 \tag{20.24} \end{align}\]

\[\begin{align} u_y &= 0 \nonumber \\ 6y -3x &= 0 \tag{20.25} \end{align}\]

\[\begin{align} u_z &= 0 \nonumber \\ 8+8z &= 0\\ z &= -1 \tag{20.26} \end{align}\]

Solving (20.24), (20.25) and (20.26), we get \(x=0, y=0, z=-1\). The stationary point is therefore \((0,0,-1)\).

At point (0,0,-1), \(u_{xx}=4>0\),

\[ \begin{vmatrix} u_{xx} & u_{xy} \\ u_{yx} & u_{yy} \end{vmatrix} = \begin{vmatrix} 4 & -3\\ -3 & 6 \end{vmatrix} =24-9=15 > 0 \]

and

\[ \begin{vmatrix} u_{xx} & u_{xy} & u_{xz} \\ u_{yx} & u_{yy} & u_{yz} \\ u_{zx} & u_{zy} & u_{zz} \end{vmatrix} = \begin{vmatrix} 4 & -3 & 0 \\ -3 & 6 & 0 \\ 0 & 0 & 8 \end{vmatrix} = 120 > 0 \]

Thus all Hessian determinants \(\begin{vmatrix} H1\end{vmatrix}, \begin{vmatrix} H2\end{vmatrix}, \begin{vmatrix} H3\end{vmatrix} >0\). Hence \(u\) has minimum value at \((0,0,-1)\) and the value is

\[\begin{equation*} \begin{split} u(0,0,-1) &= -8+4=-4 \end{split} \end{equation*}\]

22. \((x+y+z)^3 - 3(x+y+z) - 24xyz + 1\)

The given function \(u\) has three variables that can be represented as \(u(x,y,z)\).

Here,

\[\begin{equation*} \begin{split} u_x &= 3(x+y+z)^2-3-24yz\\ u_y &= 3(x+y+z)^2-3-24xz\\ u_z &= 3(x+y+z)^2-3-24xy\\ u_{xx} &= 6(x+y+z)\\ u_{xy} &= 6(x+y+z) -24z\\ u_{xz} &= 6(x+y+z)-24y\\ u_{yx} &= 6(x+y+z)-24z\\ u_{yy} &= 6(x+y+z)\\ u_{yz} &= 6(x+y+z)-24x\\ u_{zx} &= 6(x+y+z)-24y\\ u_{zy} &= 6(x+y+z)-24x\\ u_{zz} &= 6(x+y+z)\\ \end{split} \end{equation*}\]

For extreme values,

\[\begin{align} u_x &= 0 \nonumber \\ 3(x+y+z)^2-3-24yz &= 0 \tag{20.27} \end{align}\]

\[\begin{align} u_y &= 0 \nonumber \\ 3(x+y+z)^2-3-24xz &= 0 \tag{20.28} \end{align}\]

\[\begin{align} u_z &= 0 \nonumber \\ 3(x+y+z)^2-3-24xy &= 0 \tag{20.29} \end{align}\]

Solving (20.27), (20.28) and (20.29), we get \(x=y=z\). And the stationary points are \((1,1,1)\) and \((-1,-1,-1)\).

At point (1,1,1),

\[\begin{equation*} \begin{split} \begin{vmatrix}H1 \end{vmatrix} = u_{xx} &= 18 >0\\ u_{xy} &= -6\\ u_{xz} &= -6\\ u_{yx} &= -6\\ u_{yy} &= 18 \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} u_{yz} &= -6\\ u_{zx} &= -6\\ u_{zy} &= -6\\ u_{zz} &= 18 \end{split} \end{equation*}\]

\[ \begin{vmatrix}H2 \end{vmatrix}= \begin{vmatrix} u_{xx} & u_{xy} \\ u_{yx} & u_{yy} \end{vmatrix} = \begin{vmatrix} 18 & -6\\ -6 & 18 \end{vmatrix} =288 > 0 \]

and

\[ \begin{vmatrix}H3 \end{vmatrix}= \begin{vmatrix} u_{xx} & u_{xy} & u_{xz} \\ u_{yx} & u_{yy} & u_{yz} \\ u_{zx} & u_{zy} & u_{zz} \end{vmatrix} = \begin{vmatrix} 18 & -6 & -6 \\ -6 & 18 & -6 \\ -6 & -6 & 18 \end{vmatrix} = 3456 > 0 \]

Thus all Hessian determinants \(\begin{vmatrix} H1\end{vmatrix}, \begin{vmatrix} H2\end{vmatrix}, \begin{vmatrix} H3\end{vmatrix}>0\). Hence \(u\) has minimum value at \((1,1,1)\) and the value is

\[\begin{equation*} \begin{split} u(1,1,1) &= 27-9-24+1=-5 \end{split} \end{equation*}\]

At point \((-1,-1,-1)\),

\[\begin{equation*} \begin{split} \begin{vmatrix}H1 \end{vmatrix} = u_{xx} &= -18 < 0\\ u_{xy} &= 6\\ u_{xz} &= 6\\ u_{yx} &= 6 \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} u_{yy} &= -18\\ u_{yz} &= 6\\ u_{zx} &= 6\\ u_{zy} &= 6\\ u_{zz} &= -18 \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \begin{vmatrix}H2 \end{vmatrix}= \begin{vmatrix} u_{xx} & u_{xy} \\ u_{yx} & u_{yy} \end{vmatrix} = \begin{vmatrix} -18 & 6 \\ 6 & -18 \end{vmatrix} = 288 \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \begin{vmatrix}H3 \end{vmatrix}= \begin{vmatrix} u_{xx} & u_{xy} & u_{xz} \\ u_{yx} & u_{yy} & u_{yz} \\ u_{zx} & u_{zy} & u_{zz} \end{vmatrix} = \begin{vmatrix} -18 & 6 & 6 \\ 6 & -18 & 6 \\ 6 & 6 & -18 \end{vmatrix} = -3456 < 0 \end{split} \end{equation*}\]

Thus Hessian determinants \(\begin{vmatrix}H1\end{vmatrix}, \begin{vmatrix}H3\end{vmatrix} < 0\) and \(\begin{vmatrix}H2\end{vmatrix}>0\). So the function has maximum at \((-1,-1,-1)\) and the value is

\[\begin{equation*} \begin{split} u(-1,-1,-1) &= -27 + 9 + 24 + 1\\ &= 7 \end{split} \end{equation*}\]

20.0.4 Question 4

1. Find the extreme values of \(xy^2\) when \(x+y=1\).

Here, \(f(x,y)=xy^2\) and the condition or constraint given is \(x+y=1\). The function \(f\) can be expressed as the function of \(y\) alone.

\[\begin{equation*} \begin{split} f&=(1-y)y^2=y^2-y^3\\ f_y &= 2y-3y^2\\ f_{yy} &= 2-6y \end{split} \end{equation*}\]

For extreme points,

\[\begin{equation*} \begin{split} f_y &= 0\\ 2y-3y^2 &= 0\\ y(2-3y) &= 0\\ y &= 0, \frac{2}{3} \end{split} \end{equation*}\]

When \(y=0, x=1\) and when \(y=\frac{2}{3}, x=\frac{1}{3}\). So the stationary points are \((1,0)\) and \(\left(\frac{1}{3},\frac{2}{3}\right)\).

At \((1,0)\), \(f_{yy}=2 >0\). So there is minimum at this point and the value is \(f(1,0)=0\).

At \(\left(\frac{1}{3},\frac{2}{3}\right)\), \(f_{yy}=2-4=-2 <0\). So there is maximum at this point and the value is \(\frac{1}{3}\times \frac{4}{9}=\frac{4}{27}\).

2. Find the maximum value of \(u=48-(x-5)^2 - 3(y-4)^2\) if it is subjected to the condition \(x+3y=9\).

The function \(u\) can be expressed in terms of \(y\),

\[\begin{equation*} \begin{split} u &= 48-(4-3y)^2-3(y-4)^2\\ u_y &= -2(4-3y).(-3)-6(y-4)=6(4-3y)-6(y-4)\\ u_{yy} &= -24 \end{split} \end{equation*}\]

For extreme points,

\[\begin{equation*} \begin{split} u_y &= 0\\ 6(4-3y)-6(y-4) &= 0\\ 24-18y-6y+24 &= 0\\ y &= 2 \end{split} \end{equation*}\]

When \(y=2, x=3\). The stationary point is thus \((3,2)\). Here \(u_{yy}=-24<0\). So there is maxima at this point and the value is:

\[\begin{equation*} \begin{split} u(3,2) &= 48-4-12\\ &= 32 \end{split} \end{equation*}\]

3. Find the extreme value of the function \(f=x_1^2 + x_2^2 + x_3^2\) subject to the condition that \(x_1 + x_2 + x_3 = 1\).

Lets define \(F=f + \lambda \phi\) i.e. \(F=x_1^2 + x_2^2 + x_3^2+\lambda(x_1 + x_2 + x_3 - 1)\).

Taking partial derivatives,

\[\begin{equation*} \begin{split} F_{x_1} &= 2x_1 + \lambda \\ F_{x_2} &= 2x_2 + \lambda \\ F_{x_3} &= 2x_3 + \lambda \\ F_{\lambda} &= x_1 + x_2 + x_3 - 1 \\ \end{split} \end{equation*}\]

For extreme points,

\[\begin{align} F_{x_1} &= 0\\ 2x_1 + \lambda &= 0 \tag{20.30} \end{align}\]

\[\begin{align} F_{x_2} &= 0\\ 2x_2 + \lambda &= 0 \tag{20.31} \end{align}\]

\[\begin{align} F_{x_3} &= 0\\ 2x_3 + \lambda &= 0 \tag{20.32} \end{align}\]

\[\begin{align} F_{\lambda} &= 0\\ x_1 + x_2 + x_3 - 1 &= 0 \tag{20.33} \end{align}\]

Solving (20.30), (20.31), (20.32) and (20.33), we get \(x_1=x_2=x_3=\frac{1}{3}\) and \(\lambda=-\frac{2}{3}\).

Since \(f=x_1^2 + x_2^2 + x_3^2\) and \(\phi=x_1+x_2+x_3-1\),

\[\begin{equation*} \begin{split} f_{x_1} &= 2x_1\\ f_{x_2} &= 2x_2\\ f_{x_3} &= 2x_3\\ f_{x_1x_1} &= 2\\ f_{x_1x_2} &= 0\\ f_{x_1x_3} &= 0\\ f_{x_2x_1} &= 0\\ f_{x_2x_2} &= 2\\ f_{x_2x_3} &= 0\\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} f_{x_3x_1} &= 0\\ f_{x_3x_2} &= 0\\ f_{x_3x_3} &= 2\\ \phi_{x_1} &= 1\\ \phi_{x_2} &= 1\\ \phi_{x_3} &= 1\\ \end{split} \end{equation*}\]

Now, calculating bordered Hessian determinants,

\[ \begin{vmatrix} \widetilde{H1}\end{vmatrix} = \begin{vmatrix} 0 & \phi_{x_1} \\ \phi_{x_1} & f_{x_1x_1} \end{vmatrix} = \begin{vmatrix} 0 & 1\\ 1 & 2 \end{vmatrix} = -1 \]

\[ \begin{vmatrix} \widetilde{H2}\end{vmatrix} = \begin{vmatrix} 0 & \phi_{x_1} & \phi_{x_2} \\ \phi_{x_1} & f_{x_1x_1} & f_{x_1x_2} \\ \phi_{x_2} & f_{x_2x_1} & f_{x_2x_2} \end{vmatrix} = \begin{vmatrix} 0 & 1 & 1\\ 1 & 2 & 0\\ 1 & 0 & 2 \end{vmatrix} =-4 <0 \]

and

\[ \begin{vmatrix} \widetilde{H3}\end{vmatrix} = \begin{vmatrix} 0 & \phi_{x_1} & \phi_{x_2} & \phi_{x_3} \\ \phi_{x_1} & f_{x_1x_1} & f_{x_1x_2} & f_{x_1x_3} \\ \phi_{x_2} & f_{x_2x_1} & f_{x_2x_2} & f_{x_2x_3} \\ \phi_{x_3} & f_{x_3x_1} & f_{x_3x_2} & f_{x_3x_3} \end{vmatrix} =-12 <0 \]

Thus all bordered Hessian determinants \(\begin{vmatrix} \widetilde{H1}\end{vmatrix}, \begin{vmatrix} \widetilde{H2}\end{vmatrix}\) and \(\begin{vmatrix} \widetilde{H3}\end{vmatrix}\) are \(<0\). Hence \(f\) has minimum at \((\frac{1}{3},\frac{1}{3},\frac{1}{3})\) and the value is

\[\begin{equation*} \begin{split} f\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right) &= \frac{1}{9}+\frac{1}{9}+\frac{1}{9}\\ &= \frac{1}{3} \end{split} \end{equation*}\]

4. Find the minimum value of \(x^2 + y^2 + z^2\) having given \(ax+by+cz =p\).

Let \(f=x^2+y^2+z^2\) and \(\phi=ax+by+cz-p\). So

\[\begin{equation*} \begin{split} F &=f+\lambda \phi\\ &= x^2+y^2+z^2 + \lambda(ax+by+cz-p)\\ F_x &= 2x + \lambda a\\ F_y &= 2y + \lambda b\\ F_z &= 2z + \lambda c\\ F_{\lambda} &= ax+by+cz-p \end{split} \end{equation*}\]

For extreme values,

\[\begin{align} F_x &= 0\\ 2x + \lambda a &= 0 \tag{20.34} \end{align}\]

\[\begin{align} F_y &= 0\\ 2y + \lambda b &= 0 \tag{20.35} \end{align}\]

\[\begin{align} F_z &= 0\\ 2z + \lambda c &= 0 \tag{20.36} \end{align}\]

\[\begin{align} F_{\lambda} &= 0\\ ax + by + cz - p &= 0 \tag{20.37} \end{align}\]

Solving (20.34), (20.35) and (20.36), we get

\[\begin{align} \frac{x}{a} = \frac{y}{b} = \frac{z}{c} \tag{20.38} \end{align}\]

Using (20.38) in (20.37), we get

\[\begin{equation*} \begin{split} x &= \frac{ap}{a^2 + b^2 + c^2}\\ y &= \frac{bp}{a^2 + b^2 + c^2}\\ z &= \frac{cp}{a^2 + b^2 + c^2}\\ \lambda &= -\frac{2p}{a^2 + b^2 + c^2} \end{split} \end{equation*}\]

Taking partial derivatives of \(f\) and \(\phi\),

\[\begin{equation*} \begin{split} f_x & = 2x\\ f_y &= 2y\\ f_z &= 2z\\ f_{xx} &= 2\\ f_{xy} &= 0\\ f_{xz} &= 0\\ f_{yx} &= 0\\ f_{yy} &= 2\\ f_{yz} &= 0 \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} f_{zx} &= 0\\ f_{zy} &= 0\\ f_{zz} &= 2\\ \phi_x &= a\\ \phi_y &= b\\ \phi_z &= c \end{split} \end{equation*}\]

Now calculating bordered Hessian determinants,

\[ \begin{vmatrix} \widetilde{H1}\end{vmatrix} = \begin{vmatrix} 0 & \phi_{x} \\ \phi_{x} & f_{xx} \end{vmatrix} = \begin{vmatrix} 0 & a \\ a & 2 \end{vmatrix} =-a^2 <0 \]

\[ \begin{vmatrix} \widetilde{H2}\end{vmatrix} = \begin{vmatrix} 0 & \phi_{x} & \phi_{y} \\ \phi_{x} & f_{xx} & f_{xy} \\ \phi_{y} & f_{yx} & f_{yy} \end{vmatrix} = \begin{vmatrix} 0 & a & b\\ a & 2 & 0\\ b & 0 & 2 \end{vmatrix} = -2a^2 - 2b^2 <0 \]

\[ \begin{vmatrix} \widetilde{H3}\end{vmatrix} = \begin{vmatrix} 0 & \phi_{x} & \phi_{y} & \phi_{z} \\ \phi_{x} & f_{xx} & f_{xy} & f_{xz} \\ \phi_{y} & f_{yx} & f_{yy} & f_{yz} \\ \phi_{z} & f_{zx} & f_{zy} & f_{zz} \end{vmatrix} = \begin{vmatrix} 0 & a & b & c\\ a & 2 & 0 & 0 \\ b & 0 & 2 & 0\\ c & 0 & 0 & 2 \end{vmatrix} =-4a^2-4b^2-4c^2<0 \]

Thus all bordered Hessian determinants \(\begin{vmatrix} \widetilde{H1}\end{vmatrix}, \begin{vmatrix} \widetilde{H2}\end{vmatrix}\) and \(\begin{vmatrix} \widetilde{H3}\end{vmatrix}\) are \(<0\). Hence \(f\) has minimum at \(\left(\frac{ap}{a^2 + b^2 + c^2}, \frac{bp}{a^2 + b^2 + c^2},\frac{cp}{a^2 + b^2 + c^2}\right)\) and the value is

\[\begin{equation*} \begin{split} f &= x^2 + y^2 + z^2\\ &= \left(\frac{ap}{a^2 + b^2 + c^2}\right)^2 + \left(\frac{bp}{a^2 + b^2 + c^2}\right)^2 + \left(\frac{cp}{a^2 + b^2 + c^2}\right)^2\\ &= \frac{p^2}{a^2 + b^2 + c^2} \end{split} \end{equation*}\]

22. Find the minimum value of \(x^2 + y^2 + z^2\) when \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1\).

Let \(f=x^2 + y^2 + z^2\) and \(\phi=\frac{1}{x} + \frac{1}{y} + \frac{1}{z} - 1\). So

\[\begin{equation*} \begin{split} F &=f+\lambda \phi\\ &= x^2+y^2+z^2 + \lambda\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1\right)\\ F_x &= 2x-\frac{\lambda}{x^2} \\ F_y &= 2y-\frac{\lambda}{y^2} \\ F_z &= 2z-\frac{\lambda}{z^2} \\ F_{\lambda} &= \frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1 \end{split} \end{equation*}\]

For extreme values,

\[\begin{align} F_x &= 0\\ 2x-\frac{\lambda}{x^2} &= 0 \tag{20.39} \end{align}\]

\[\begin{align} F_y &= 0\\ 2y-\frac{\lambda}{y^2} &= 0 \tag{20.40} \end{align}\]

\[\begin{align} F_z &= 0\\ 2z-\frac{\lambda}{z^2} &= 0 \tag{20.41} \end{align}\]

\[\begin{align} F_{\lambda} &= 0\\ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1 &= 0 \tag{20.42} \end{align}\]

Solving (20.39), (20.40) and (20.41), we get \(x=y=z\).

Putting this into (20.42), we get \(x=y=z=3\).

Taking partial derivatives of \(f\) and \(\phi\),

\[\begin{equation*} \begin{split} f_x & = 2x\\ f_y &= 2y\\ f_z &= 2z\\ f_{xx} &= 2\\ f_{xy} &= 0\\ f_{xz} &= 0\\ f_{yx} &= 0\\ f_{yy} &= 2\\ f_{yz} &= 0 \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} f_{zx} &= 0\\ f_{zy} &= 0\\ f_{zz} &= 2\\ \phi_x &= -\frac{1}{x^2}=-\frac{1}{9}\\ \phi_y &= -\frac{1}{y^2}=-\frac{1}{9}\\ \phi_z &= -\frac{1}{z^2}=-\frac{1}{9} \end{split} \end{equation*}\]

Now calculating bordered Hessian determinants,

\[ \begin{vmatrix} \widetilde{H1}\end{vmatrix} = \begin{vmatrix} 0 & \phi_{x} \\ \phi_{x} & f_{xx} \end{vmatrix} = \begin{vmatrix} 0 & -\frac{1}{9} \\ -\frac{1}{9} & 2 \end{vmatrix} = -\frac{1}{81} <0 \]

\[ \begin{vmatrix} \widetilde{H2}\end{vmatrix} = \begin{vmatrix} 0 & \phi_{x} & \phi_{y} \\ \phi_{x} & f_{xx} & f_{xy} \\ \phi_{y} & f_{yx} & f_{yy} \end{vmatrix} = \begin{vmatrix} 0 & -\frac{1}{9} & -\frac{1}{9} \\ -\frac{1}{9} & 2 & 0 \\ -\frac{1}{9} & 0 & 2 \end{vmatrix} =-\frac{4}{81} <0 \]

\[ \begin{vmatrix} \widetilde{H3}\end{vmatrix} = \begin{vmatrix} 0 & \phi_{x} & \phi_{y} & \phi_{z} \\ \phi_{x} & f_{xx} & f_{xy} & f_{xz} \\ \phi_{y} & f_{yx} & f_{yy} & f_{yz} \\ \phi_{z} & f_{zx} & f_{zy} & f_{zz} \end{vmatrix} = \begin{vmatrix} 0 & -\frac{1}{9} & -\frac{1}{9} & -\frac{1}{9} \\ -\frac{1}{9} & 2 & 0 & 0\\ -\frac{1}{9} & 0 & 2 & 0\\ -\frac{1}{9} & 0 & 0 & 2 \end{vmatrix} = - 0.148 <0 \]

Thus all bordered Hessian determinants \(\begin{vmatrix} \widetilde{H1}\end{vmatrix} , \begin{vmatrix} \widetilde{H2}\end{vmatrix}\) and \(\begin{vmatrix} \widetilde{H3}\end{vmatrix}\) are \(<0\). Hence \(f\) has minimum at \((3,3,3)\) and the value is

\[\begin{equation*} \begin{split} f(3,3,3) &= 27 \end{split} \end{equation*}\]

20.1 Useful links