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=x2−xy+y2+3x−2y+1$f=x^2-xy+y^2+3x-2y+1$

Here,

fx=2x−y+3fy=−x+2y−2fxx=2fxy=−1fyy=2$$\begin{array}{cc}{f}_x& =2x-y+3\\ f_y& =-x+2y-2\\ f_{xx}& =2\\ f_{xy}& =-1\\ f_{yy}& =2\end{array}$$

For extreme values,

fx=02x−y+3=0$$\begin{array}{cc}{f}_x& =0\\ 2x-y+3& =0& \text{(20.1)}\end{array}$$

fy=0−x+2y−2=0$$\begin{array}{cc}{f}_y& =0\\ -x+2y-2& =0& \text{(20.2)}\end{array}$$

Solving (20.1) and (20.2), we get x=−43,y=13$x=-\frac{4}{3},y=\frac{1}{3}$.

For point (−43,13)$(-\frac{4}{3},\frac{1}{3})$,

fxy=−1$$\begin{array}{cc}{f}_{xy}& =-1\end{array}$$

Thus fxx>0$f_{xx}>0$ and fxxfyy−(fxy)2=2×2−(−1)2=3$f_{xx}{f}_{yy}-(f_{xy}{)}^2=2\times 2-(-1{)}^2=3$ which is >0$>0$. Thus f(x,y)$f(x,y)$ has a minimum at (−43,13)$(-\frac{4}{3},\frac{1}{3})$. Its value is

=169+49+19−4−23+1=219−(9+23)=219−113=−43$$\begin{array}{cc}& =\frac{16}{9}+\frac{4}{9}+\frac{1}{9}-4-\frac{2}{3}+1\\ & =\frac{21}{9}-\left(\frac{9+2}{3}\right)\\ & =\frac{21}{9}-\frac{11}{3}\\ & =-\frac{4}{3}\end{array}$$

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2. u=16−(x+2)2−(y−2)2$u=16-(x+2{)}^2-(y-2{)}^2$

Here,

ux=−2(x+2)uy=−2(y−2)uxx=−2uxy=0uyy=−2$$\begin{array}{cc}{u}_x& =-2(x+2)\\ u_y& =-2(y-2)\\ u_{xx}& =-2\\ u_{xy}& =0\\ u_{yy}& =-2\end{array}$$

For extreme values,

ux=0x=−2$$\begin{array}{cc}{u}_x& =0\\ x& =-2\end{array}$$

uy=0y=2$$\begin{array}{cc}{u}_y& =0\\ y& =2\end{array}$$

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

uxxuyy−(uxy)2=(−2)×(−2)−0=4>0$$\begin{array}{cc}{u}_{xx}{u}_{yy}-(u_{xy}{)}^2& =(-2)\times (-2)-0\\ & =4>0\end{array}$$

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

=16−0−0=16$$\begin{array}{cc}& =16-0-0\\ & =16\end{array}$$

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3. z=8−4x+4y−x2−y2$z=8-4x+4y-x^2-y^2$

Here,

zx=−4−2xzy=4−2yzxx=−2zxy=0zyy=−2$$\begin{array}{cc}{z}_x& =-4-2x\\ z_y& =4-2y\\ z_{xx}& =-2\\ z_{xy}& =0\\ z_{yy}& =-2\end{array}$$

For extreme values,

zx=0x=−2$$\begin{array}{cc}{z}_x& =0\\ x& =-2\end{array}$$

zy=0y=2$$\begin{array}{cc}{z}_y& =0\\ y& =2\end{array}$$

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

zxxzyy−(zxy)2=(−2)×(−2)−0=4>0$$\begin{array}{cc}{z}_{xx}{z}_{yy}-(z_{xy}{)}^2& =(-2)\times (-2)-0\\ & =4>0\end{array}$$

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

=8+8+8−4−4=16$$\begin{array}{cc}& =8+8+8-4-4\\ & =16\end{array}$$

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4. z=x3−x2−y2+xy$z=x^3-x^2-y^2+xy$

Here,

zx=3x2−2x+yzy=−2y+xzxx=6x−2zxy=1zyy=−2$$\begin{array}{cc}{z}_x& =3x^2-2x+y\\ z_y& =-2y+x\\ z_{xx}& =6x-2\\ z_{xy}& =1\\ z_{yy}& =-2\end{array}$$

For extreme values,

zx=03x2−2x+y=0$$\begin{array}{cc}{z}_x& =0\\ 3x^2-2x+y& =0& \text{(20.3)}\end{array}$$

zy=0x−2y=0$$\begin{array}{cc}{z}_y& =0\\ x-2y& =0& \text{(20.4)}\end{array}$$

Solving (20.3) and (20.4), we get x=0,12$x=0,\frac{1}{2}$.

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

At (0,0)$(0,0)$,

zxx=−2zxxzyy−(zxy)2=(−2)×(−2)−1=3$$\begin{array}{cc}{z}_{xx}& =-2\\ z_{xx}{z}_{yy}-(z_{xy}{)}^2& =(-2)\times (-2)-1\\ & =3\end{array}$$

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

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

zxx=1zxxzyy−(zxy)2=1×(−2)−1=−3$$\begin{array}{cc}{z}_{xx}& =1\\ z_{xx}{z}_{yy}-(z_{xy}{)}^2& =1\times (-2)-1\\ & =-3\end{array}$$

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

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22. u=2x2+xy+4y2+xz+z2+2$u=2x^2+xy+4y^2+xz+z^2+2$

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

Here,

ux=4x+y+zuy=x+8yuz=x+2zuxx=4uxy=1uxz=1uyx=1uyy=8uyz=0uzx=1uzy=0uzz=2$$\begin{array}{cc}{u}_x& =4x+y+z\\ u_y& =x+8y\\ u_z& =x+2z\\ u_{xx}& =4\\ u_{xy}& =1\\ u_{xz}& =1\\ u_{yx}& =1\\ u_{yy}& =8\\ u_{yz}& =0\\ u_{zx}& =1\\ u_{zy}& =0\\ u_{zz}& =2\end{array}$$

For extreme values,

ux=04x+y+z=0$$\begin{array}{cc}{u}_x& =0\\ 4x+y+z& =0& \text{(20.5)}\end{array}$$

uy=0x+8y=0$$\begin{array}{cc}{u}_y& =0\\ x+8y& =0& \text{(20.6)}\end{array}$$

uz=0x+2z=0$$\begin{array}{cc}{u}_z& =0\\ x+2z& =0& \text{(20.7)}\end{array}$$

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

Here, uxx>0$u_{xx}>0$,

|uxxuxyuyxuyy|=|4118|=31>0$$\left|\begin{array}{cc}{u}_{xx}& u_{xy}\\ u_{yx}& u_{yy}\end{array}\right|=\left|\begin{array}{cc}4& 1\\ 1& 8\end{array}\right|=31>0$$

and

|uxxuxyuxzuyxuyyuyzuzxuzyuzz|=|411180102|=54>0$$\left|\begin{array}{ccc}{u}_{xx}& u_{xy}& u_{xz}\\ u_{yx}& u_{yy}& u_{yz}\\ u_{zx}& u_{zy}& u_{zz}\end{array}\right|=\left|\begin{array}{ccc}4& 1& 1\\ 1& 8& 0\\ 1& 0& 2\end{array}\right|=54>0$$

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

u(0,0,0)=2$$\begin{array}{cc}u(0,0,0)& =2\end{array}$$

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6. ϕ=35−(2x+3)2−(y−4)2−(z+1)2$\varphi =35-(2x+3{)}^2-(y-4{)}^2-(z+1{)}^2$

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

Here,

ϕx=−2(2x+3)×2=−4(2x+3)ϕy=−2(y−4)ϕz=−2(z+1)ϕxx=−8ϕxy=0ϕxz=0ϕyx=0ϕyy=−2ϕyz=0ϕzx=0ϕzy=0ϕzz=−2$$\begin{array}{cc}{\varphi }_x& =-2(2x+3)\times 2=-4(2x+3)\\ {\varphi }_y& =-2(y-4)\\ {\varphi }_z& =-2(z+1)\\ {\varphi }_{xx}& =-8\\ {\varphi }_{xy}& =0\\ {\varphi }_{xz}& =0\\ {\varphi }_{yx}& =0\\ {\varphi }_{yy}& =-2\\ {\varphi }_{yz}& =0\\ {\varphi }_{zx}& =0\\ {\varphi }_{zy}& =0\\ {\varphi }_{zz}& =-2\end{array}$$

For extreme values,

ϕx=0−4(2x+3)=0x=−32$$\begin{array}{cc}{\varphi }_x& =0\\ -4(2x+3)& =0\\ x& =-\frac{3}{2}\end{array}$$

ϕy=0−2(y−4)=0y=4$$\begin{array}{cc}{\varphi }_y& =0\\ -2(y-4)& =0\\ y& =4\end{array}$$

ϕz=0−2(z+1)=0z=−1$$\begin{array}{cc}{\varphi }_z& =0\\ -2(z+1)& =0\\ z& =-1\end{array}$$

The stationary point is thus (−32,4,−1)$(-\frac{3}{2},4,-1)$.

Here, ϕxx=−8<0${\varphi }_{xx}=-8<0$,

|ϕxxϕxyϕyxϕyy|=|−800−2|=16>0$$\left|\begin{array}{cc}{\varphi }_{xx}& {\varphi }_{xy}\\ {\varphi }_{yx}& {\varphi }_{yy}\end{array}\right|=\left|\begin{array}{cc}-8& 0\\ 0& -2\end{array}\right|=16>0$$

and

|ϕxxϕxyϕxzϕyxϕyyϕyzϕzxϕzyϕzz|=|−8000−2000−2|=−32<0$$\left|\begin{array}{ccc}{\varphi }_{xx}& {\varphi }_{xy}& {\varphi }_{xz}\\ {\varphi }_{yx}& {\varphi }_{yy}& {\varphi }_{yz}\\ {\varphi }_{zx}& {\varphi }_{zy}& {\varphi }_{zz}\end{array}\right|=\left|\begin{array}{ccc}-8& 0& 0\\ 0& -2& 0\\ 0& 0& -2\end{array}\right|=-32<0$$

Thus ϕ$\varphi$ has maximum value at (−3/2,4,−1)$(-3/2,4,-1)$ and the value is

ϕ(−32,4,−1)=35−0−0−0=35$$\begin{array}{ccc}\varphi (-\frac{3}{2},4,-1)& =35-0-0-0& =35\end{array}$$

20.0.2 Question 2

Examine for maxima and minima of following functions

1. 2x2−3xy+4y2+5$2x^2-3xy+4y^2+5$ (TU 2065)

Let u=2x2−3xy+4y2+5$u=2x^2-3xy+4y^2+5$, this is a function of two variables u(x,y)$u(x,y)$.

ux=4x−3yuy=−3x+8yuxx=4uxy=−3uyy=8uyx=−3$$\begin{array}{cc}{u}_x& =4x-3y\\ u_y& =-3x+8y\\ u_{xx}& =4\\ u_{xy}& =-3\\ u_{yy}& =8\\ u_{yx}& =-3\end{array}$$

For extreme values,

ux=04x−3y=0$$\begin{array}{cc}{u}_x& =0\\ 4x-3y& =0& \text{(20.8)}\end{array}$$

uy=0−3x+8y=0$$\begin{array}{cc}{u}_y& =0\\ -3x+8y& =0& \text{(20.9)}\end{array}$$

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

Here, uxx=4>0$u_{xx}=4>0$ and

|uxxuxyuyxuyy|=|4−3−38|=32−9=23>0$$\left|\begin{array}{cc}{u}_{xx}& u_{xy}\\ u_{yx}& u_{yy}\end{array}\right|=\left|\begin{array}{cc}4& -3\\ -3& 8\end{array}\right|=32-9=23>0$$

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

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2. xy+a3x+a3y$xy+\frac{a^3}{x}+\frac{a^3}{y}$

Let f=xy+a3x+a3y$f=xy+\frac{a^3}{x}+\frac{a^3}{y}$ which is a function of two variables f(x,y)$f(x,y)$.

fx=y−a3x2fy=x−a3y2fxx=2a3x3fyy=2a3y3$$\begin{array}{cc}{f}_x& =y-\frac{a^3}{x^2}\\ f_y& =x-\frac{a^3}{y^2}\\ f_{xx}& =\frac{2a^3}{x^3}\\ f_{yy}& =\frac{2a^3}{y^3}\end{array}$$

For extreme values,

fx=0y−a3x2=0x2y=a3$$\begin{array}{cc}{f}_x& =0\\ y-\frac{a^3}{x^2}& =0\\ x^{2}y& =a^3& \text{(20.10)}\end{array}$$

fy=0x−a3y2=0xy2=a3$$\begin{array}{cc}{f}_y& =0\\ x-\frac{a^3}{y^2}& =0\\ x{y}^2& =a^3& \text{(20.11)}\end{array}$$

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

|fxxfxyfyxfyy|=|2112|=3>0$$\left|\begin{array}{cc}{f}_{xx}& f_{xy}\\ f_{yx}& f_{yy}\end{array}\right|=\left|\begin{array}{cc}2& 1\\ 1& 2\end{array}\right|=3>0$$

Thus function has minimum at (a,a)$(a,a)$ and the value is a2+a2+a2=3a2$a^2+a^2+a^2=3a^2$.

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3. x3+y3−12x−3y+20$x^3+y^3-12x-3y+20$

Let f=x3+y3−12x−3y+20$f=x^3+y^3-12x-3y+20$ which is a function of two variables f(x,y)$f(x,y)$.

fx=3x2−12fy=3y2−3fxx=6xfyy=6yfxy=0fyx=0$$\begin{array}{cc}{f}_x& =3x^2-12\\ f_y& =3y^2-3\\ f_{xx}& =6x\\ f_{yy}& =6y\\ f_{xy}& =0\\ f_{yx}& =0\end{array}$$

For extreme values,

fx=03x2−12=0x=±2fy=03y2−3=0y=±1$$\begin{array}{cc}{f}_x& =0\\ 3x^2-12& =0\\ x& =\pm 2\\ f_y& =0\\ 3y^2-3& =0\\ y& =\pm 1\end{array}$$

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

At (2,1)$(2,1)$,

fxx=12>0fyy=6$$\begin{array}{cc}{f}_{xx}& =12>0\\ f_{yy}& =6\end{array}$$

|fxxfxyfyxfyy|=|12006|=72>0$$\left|\begin{array}{cc}{f}_{xx}& f_{xy}\\ f_{yx}& f_{yy}\end{array}\right|=\left|\begin{array}{cc}12& 0\\ 0& 6\end{array}\right|=72>0$$

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

At (2,−1)$(2,-1)$,

fxx=12>0fyy=−6$$\begin{array}{cc}{f}_{xx}& =12>0\\ f_{yy}& =-6\end{array}$$

and

|fxxfxyfyxfyy|=|1200−6|=−72<0$$\left|\begin{array}{cc}{f}_{xx}& f_{xy}\\ f_{yx}& f_{yy}\end{array}\right|=\left|\begin{array}{cc}12& 0\\ 0& -6\end{array}\right|=-72<0$$

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

At (−2,1)$(-2,1)$,

fxx=−12<0fyy=6$$\begin{array}{cc}{f}_{xx}& =-12<0\\ f_{yy}& =6\end{array}$$

and

|fxxfxyfyxfyy|=|−12006|=−72<0$$\left|\begin{array}{cc}{f}_{xx}& f_{xy}\\ f_{yx}& f_{yy}\end{array}\right|=\left|\begin{array}{cc}-12& 0\\ 0& 6\end{array}\right|=-72<0$$

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

At (−2,−1)$(-2,-1)$,

fxx=−12<0fyy=−6$$\begin{array}{cc}{f}_{xx}& =-12<0\\ f_{yy}& =-6\end{array}$$

and

|fxxfxyfyxfyy|=|−1200−6|=72>0$$\left|\begin{array}{cc}{f}_{xx}& f_{xy}\\ f_{yx}& f_{yy}\end{array}\right|=\left|\begin{array}{cc}-12& 0\\ 0& -6\end{array}\right|=72>0$$

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

Figure 20.1: Plotting of x3+y3−12x−3y+20$x^3+y^3-12x-3y+20$ showing two saddle points, one maximum and one minimum

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4. y2+2x2−5x4+4x5$y^2+2x^2-5x^4+4x^5$

Let f=y2+2x2−5x4+4x5$f=y^2+2x^2-5x^4+4x^5$ which is a function of two variables (x,y)$(x,y)$.

fx=4x−20x3+20x4fy=2yfxx=4−60x2+80x3fxy=0fyx=0fyy=2$$\begin{array}{cc}{f}_x& =4x-20x^3+20x^4\\ f_y& =2y\\ f_{xx}& =4-60x^2+80x^3\\ f_{xy}& =0\\ f_{yx}& =0\\ f_{yy}& =2\end{array}$$

For extreme values ,

fx=04x−20x3+20x4=0x(1−5x2+5x3)=0$$\begin{array}{cc}{f}_x& =0\\ 4x-20x^3+20x^4& =0\\ x(1-5x^2+5x^3)& =0& \text{(20.12)}\end{array}$$

fy=02y=0y=0$$\begin{array}{cc}{f}_y& =0\\ 2y& =0\\ y& =0& \text{(20.13)}\end{array}$$

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

Real solutions are x=0,−3√3√2110+17103−133√3√2110+1710+13$x=0,-\frac{\sqrt[3]{3\frac{3\sqrt{21}}{10}+\frac{17}{10}}}{3}-\frac{1}{3\sqrt[3]{3\frac{3\sqrt{21}}{10}+\frac{17}{10}}}+\frac{1}{3}$

Imaginary solutions are x=13−13(−12−√3i2)3√3√2110+1710−(−12−√3i2)3√3√2110+17103,13−(−12+√3i2)3√3√2110+17103−13(−12+√3i2)3√3√2110+1710

Solving 1−5x2+5x3 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),

fxx=4>0fyy=2fxy=0fyx=0

and

|fxxfxyfyxfyy|=|4002|=8>0

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

Figure 20.2: Plotting of y2+2x2−5x4+4x5

Plotting shows minimum at (0,0).

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22. −3x2+6xz+4y−2y2−6z2

The given function has three variables. Let u(x,y,z)=−3x2+6xz+4y−2y2−6z2.

Here,

ux=−6x+6zuy=4−4yuz=6x−12zuxx=−6uxy=0uxz=6uyx=0uyy=−4uyz=0uzx=6uzy=0uzz=−12

For extreme values,

ux=0−6x+6z=0−x+z=0

uy=04−4y=01−y=0

uz=06x−12z=0x−2z=0

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, uxx=−6<0,

|uxxuxyuyxuyy|=|−600−4|=24>0

and

|uxxuxyuxzuyxuyyuyzuzxuzyuzz|=|−6060−4060−12|=−144<0

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

u(0,1,0)=4−2=2

20.0.3 Question 3

Find the maximum and minimum values of

1. x3y2(1−x−y),x≠0,y≠0,x+y≠1

Here f=x3y2(1−x−y),x≠0,y≠0,x+y≠1.

fx=y2[x3.(−1)+(1−x−y).3x2]=−x3y2+3x2y2(1−x−y)fyx=−2x3y+3x2[y2.(−1)+(1−x−y).2y]=−2x3y−3x2y2+6x2y(1−x−y)fy=x3[y2.(−1)+(1−x−y).2y]=−x3y2+2x3y(1−x−y)fxy=−3x2y2+2y[x3.(−1)+(1−x−y).(3x2)]=−3x2y2−2x3y+6x2y(1−x−y)=−2x3y−3x2y2+6x2y(1−x−y)fxx=−3x2y2+3y2[x2.(−1)+(1−x−y).(2x)]=−6x2y2+6xy2(1−x−y)fyy=−2x3y+2x3[y.(−1)+(1−x−y)]=−4x3y+2x3(1−x−y)

For extreme values,

fx=0−x3y2+3x2y2(1−x−y)=0

fy=0−x3y2+2x3y(1−x−y)=0

Solving (20.17) and (20.18),

(1−x−y)(3x2y2−2x3y)=0x2y(1−x−y)(3y−2x)=0

As x≠0,y≠0,x+y≠1,

3y−2x=0x=32y

Substituting this in (20.17),

−278y3y2+3×94y2y2(1−32y−y)=0y4(−278y+274(1−5y2))=0y4(y(−278×6)+274)=0

As y≠0,

y(−278×6)+274=0y=−274×8−27×6y=13

Plugging this into (20.19),

x=12

The stationary point is thus (12,13). At this point we have,

fxx=−6×14×19+6×12×19(1−12−13)=−16+136−3−26=−19fyy=−4×18×13+2×18(1−12−13)=−16+14×16=−18fxy=−2×18×13−3×14×19+6×14×13(1−12−13)=−112−112+12×16=−112

Here fxx=−19<0,

|fxxfxyfyxfyy|=|−19−112−112−18|=172−1144=1144>0

So f has maximum at (12,13) and the value is

f(12,13)=18×19(1−12−13)=18×19×16=1432

2. 2(x−y)2−x4−y4

Let f=2(x−y)2−x4−y4 which is a function with two variables x,y. Taking partial derivatives,

fx=4(x−y)−4x3fy=−4(x−y)−4y3fxx=4−12x2fyy=4−12y2fxy=−4fyx=−4

For extreme values,

fx=04(x−y)−4x3=0

fy=0−4(x−y)−4y3=0

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

4×(−2y)+4y3=0y3−2y=0y(y2−2)=0y=0,+√2,−√2

So the points with extreme values are (0,0),(−√2,√2) and (√2,−√2).

At (0,0),

fxx=4>0fyy=4fxy,fyx=−4

and

|fxxfxyfyxfyy|=|4−4−44|=16−16=0

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

At (−√2,√2),

fxx=−20<0fyy=−20fxy,fyx=−4

and

|fxxfxyfyxfyy|=|−20−4−4−20|=384>0

So f has maximum at (−√2,√2).

At (√2,−√2),

fxx=−20<0fyy=−20fxy,fyx=−4

and

|fxxfxyfyxfyy|=|−20−4−4−20|=384>0

So f has maximum at (√2,−√2).

The function has thus maximum at two points (−√2,√2) and (√2,−√2) and the value is 8 at both points.

The plotting indeed proves this.

Figure 20.3: Plotting of 2(x−y)2−x4−y4

3. x3+3xy2−15x2−15y2+72x

This is a function of two variables f(x,y)=x3+3xy2−15x2−15y2+72x.

fx=3x2+3y2−30x+72fy=6xy−30yfxx=6x−30fyy=6x−30fxy=6yfyx=6y

For extreme values,

fx=03x2+3y2−30x+72=0x2+y2−10x+24=0

fy=06xy−30y=0y(x−5)=0

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

At (4,0),

fxx=−6<0fyy=−6fxy=0fyx=0

|fxxfxyfyxfyy| = |−600−6|

=36 > 0

So there is maximum at this point and the value is 64−240+288=112.

At (6,0),

fxx=6>0fyy=6fxy=0fyx=0

|fxxfxyfyxfyy| = |6006|

=36 > 0

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

At (5,1),

fxx=0fyy=0fxy=6fyx=6

|fxxfxyfyxfyy| = |0660|

=-36 < 0

At (5,−1),

fxx=0fyy=0fxy=−6fyx=−6

|fxxfxyfyxfyy| = |0−6−60|

=-36 < 0

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

Figure 20.4: Plotting of x3+3xy2−15x2−15y2+72x

4. 8z+2x2+3y2+4z2−3xy

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

Here,

ux=4x−3yuy=6y−3xuz=8+8zuxx=4uxy=−3uxz=0uyx=−3uyy=6uyz=0uzx=0uzy=0uzz=8

For extreme values,

ux=04x−3y=0

uy=06y−3x=0

uz=08+8z=0z=−1

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), uxx=4>0,

|uxxuxyuyxuyy|=|4−3−36|=24−9=15>0

and

|uxxuxyuxzuyxuyyuyzuzxuzyuzz|=|4−30−360008|=120>0

Thus all Hessian determinants |H1|,|H2|,|H3|>0. Hence u has minimum value at (0,0,−1) and the value is

u(0,0,−1)=−8+4=−4

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,

ux=3(x+y+z)2−3−24yzuy=3(x+y+z)2−3−24xzuz=3(x+y+z)2−3−24xyuxx=6(x+y+z)uxy=6(x+y+z)−24zuxz=6(x+y+z)−24yuyx=6(x+y+z)−24zuyy=6(x+y+z)uyz=6(x+y+z)−24xuzx=6(x+y+z)−24yuzy=6(x+y+z)−24xuzz=6(x+y+z)

For extreme values,

ux=03(x+y+z)2−3−24yz=0

uy=03(x+y+z)2−3−24xz=0

uz=03(x+y+z)2−3−24xy=0

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),

|H1|=uxx=18>0uxy=−6uxz=−6uyx=−6uyy=18

uyz=−6uzx=−6uzy=−6uzz=18

|H2|=|uxxuxyuyxuyy|=|18−6−618|=288>0

and

|H3|=|uxxuxyuxzuyxuyyuyzuzxuzyuzz|=|18−6−6−618−6−6−618|=3456>0

Thus all Hessian determinants |H1|,|H2|,|H3|>0. Hence u has minimum value at (1,1,1) and the value is

u(1,1,1)=27−9−24+1=−5

At point (−1,−1,−1),

|H1|=uxx=−18<0uxy=6uxz=6uyx=6

uyy=−18uyz=6uzx=6uzy=6uzz=−18

|H2|=|uxxuxyuyxuyy|=|−1866−18|=288

|H3|=|uxxuxyuxzuyxuyyuyzuzxuzyuzz|=|−18666−18666−18|=−3456<0

Thus Hessian determinants |H1|,|H3|<0 and |H2|>0. So the function has maximum at (−1,−1,−1) and the value is

u(−1,−1,−1)=−27+9+24+1=7

20.0.4 Question 4

1. Find the extreme values of xy2 when x+y=1.

Here, f(x,y)=xy2 and the condition or constraint given is x+y=1. The function f can be expressed as the function of y alone.

f=(1−y)y2=y2−y3fy=2y−3y2fyy=2−6y

For extreme points,

fy=02y−3y2=0y(2−3y)=0y=0,23

When y=0,x=1 and when y=23,x=13. So the stationary points are (1,0) and (13,23).

At (1,0), fyy=2>0. So there is minimum at this point and the value is f(1,0)=0.

At (13,23), fyy=2−4=−2<0. So there is maximum at this point and the value is 13×49=427.

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,

u=48−(4−3y)2−3(y−4)2uy=−2(4−3y).(−3)−6(y−4)=6(4−3y)−6(y−4)uyy=−24

For extreme points,

uy=06(4−3y)−6(y−4)=024−18y−6y+24=0y=2

When y=2,x=3. The stationary point is thus (3,2). Here uyy=−24<0. So there is maxima at this point and the value is:

u(3,2)=48−4−12=32

3. Find the extreme value of the function f=x21+x22+x23 subject to the condition that x1+x2+x3=1.

Lets define F=f+λϕ i.e. F=x21+x22+x23+λ(x1+x2+x3−1).

Taking partial derivatives,

Fx1=2x1+λFx2=2x2+λFx3=2x3+λFλ=x1+x2+x3−1

For extreme points,

Fx1=02x1+λ=0

Fx2=02x2+λ=0

Fx3=02x3+λ=0

Fλ=0x1+x2+x3−1=0

Solving (20.30), (20.31), (20.32) and (20.33), we get x1=x2=x3=13 and λ=−23.

Since f=x21+x22+x23 and ϕ=x1+x2+x3−1,

fx1=2x1fx2=2x2fx3=2x3fx1x1=2fx1x2=0fx1x3=0fx2x1=0fx2x2=2fx2x3=0

fx3x1=0fx3x2=0fx3x3=2ϕx1=1ϕx2=1ϕx3=1

Now, calculating bordered Hessian determinants,

|~H1|=|0ϕx1ϕx1fx1x1|=|0112|=−1

|~H2|=|0ϕx1ϕx2ϕx1fx1x1fx1x2ϕx2fx2x1fx2x2|=|011120102|=−4<0

and

|~H3|=|0ϕx1ϕx2ϕx3ϕx1fx1x1fx1x2fx1x3ϕx2fx2x1fx2x2fx2x3ϕx3fx3x1fx3x2fx3x3|=−12<0

Thus all bordered Hessian determinants |~H1|,|~H2| and |~H3| are <0. Hence f has minimum at (13,13,13) and the value is

f(13,13,13)=19+19+19=13

4. Find the minimum value of x2+y2+z2 having given ax+by+cz=p.

Let f=x2+y2+z2 and ϕ=ax+by+cz−p. So

F=f+λϕ=x2+y2+z2+λ(ax+by+cz−p)Fx=2x+λaFy=2y+λbFz=2z+λcFλ=ax+by+cz−p

For extreme values,

Fx=02x+λa=0

Fy=02y+λb=0

Fz=02z+λc=0

Fλ=0ax+by+cz−p=0

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

xa=yb=zc

Using (20.38) in (20.37), we get

x=apa2+b2+c2y=bpa2+b2+c2z=cpa2+b2+c2λ=−2pa2+b2+c2

Taking partial derivatives of f and ϕ,

fx=2xfy=2yfz=2zfxx=2fxy=0fxz=0fyx=0fyy=2fyz=0

fzx=0fzy=0fzz=2ϕx=aϕy=bϕz=c

Now calculating bordered Hessian determinants,

|~H1|=|0ϕxϕxfxx|=|0aa2|=−a2<0

|~H2|=|0ϕxϕyϕxfxxfxyϕyfyxfyy|=|0aba20b02|=−2a2−2b2<0

|~H3|=|0ϕxϕyϕzϕxfxxfxyfxzϕyfyxfyyfyzϕzfzxfzyfzz|=|0abca200b020c002|=−4a2−4b2−4c2<0

Thus all bordered Hessian determinants |~H1|,|~H2| and |~H3| are <0. Hence f has minimum at (apa2+b2+c2,bpa2+b2+c2,cpa2+b2+c2) and the value is

f=x2+y2+z2=(apa2+b2+c2)2+(bpa2+b2+c2)2+(cpa2+b2+c2)2=p2a2+b2+c2

22. Find the minimum value of x2+y2+z2 when 1x+1y+1z=1.

Let f=x2+y2+z2 and ϕ=1x+1y+1z−1. So

F=f+λϕ=x2+y2+z2+λ(1x+1y+1z−1)Fx=2x−λx2Fy=2y−λy2Fz=2z−λz2Fλ=1x+1y+1z−1

For extreme values,

Fx=02x−λx2=0

Fy=02y−λy2=0

Fz=02z−λz2=0

Fλ=01x+1y+1z−1=0

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 ϕ,

fx=2xfy=2yfz=2zfxx=2fxy=0fxz=0fyx=0fyy=2fyz=0

fzx=0fzy=0fzz=2ϕx=−1x2=−19ϕy=−1y2=−19ϕz=−1z2=−19

Now calculating bordered Hessian determinants,

|~H1|=|0ϕxϕxfxx|=|0−19−192|=−181<0

|~H2|=|0ϕxϕyϕxfxxfxyϕyfyxfyy|=|0−19−19−1920−1902|=−481<0

|~H3|=|0ϕxϕyϕzϕxfxxfxyfxzϕyfyxfyyfyzϕzfzxfzyfzz|=|0−19−19−19−19200−19020−19002|=−0.148<0

Thus all bordered Hessian determinants |~H1|,|~H2| and |~H3| are <0. Hence f has minimum at (3,3,3) and the value is

f(3,3,3)=27

20.1 Useful links