Q 20. - Calculus

Question

Consider a function of two variables, $f(x,y)=x^m{y}^n$ where m and n are positive integers. What conditions do m and n have to satisfy in order for f to have a relative minimum at the origin? What conditions do m and n have to satisfy in order for f to have a saddle point at the origin? Are there any other possibilities for the behavior of the function at the origin? It may be helpful to

refer to your answers to Exercise 19 before you answer this question 

Step-by-Step Solution

Answer

The condition for saddle point is $m+n>1$

Step 1: Given information

$f(x,y)=x^m{y}^n$

Step 2: Calculation

Consider the function

$f(x,y)=x^m{y}^n\dots \dots .\left(1\right)$

where $m$ and $n$ are integers.

The goal is to discover the condition where the function $f(x, y)$ has a minimum at the origin for $m$ and $n$

Partially differentiate $\left(1\right)$ with regard to $x$

$$\begin{gathered} \frac{\partial }{\partial x}f(x,y)=\frac{\partial }{\partial x}\left\{x^m{y}^n\right\} \\ {f}_x(x,y)=y^n\frac{\partial }{\partial x}{x}^m \\ =m{y}^n{x}^{m-1}\cdots \cdots \left(2\right) \end{gathered}$$

Partially differentiate $\left(1\right)$ with regard to $y$

$$\begin{gathered} \frac{\partial }{\partial y}f(x,y)=\frac{\partial }{\partial y}\left\{x^m{y}^n\right\} \\ {f}_y(x,y)=x^n\frac{\partial }{\partial y}{y}^n \end{gathered}$$

$=n{x}^m{y}^{n-1}\dots \dots \left(3\right)$

The stationary point is glven by

$f_x(x,y)=0$ and $f_y(x,y)=0$

Thus,

$$\begin{gathered} f_x(x,y)=0 \\ m{y}^n{x}^{m-1}=0 \end{gathered}$$

Either, $x=0$ or $y=0$

And,

$$\begin{gathered} f_y(x,y)=0 \\ n{x}^m{y}^{n-1}=0 \end{gathered}$$

Either, $x=0$ or $y=0$

Step 3: Calculation

Calculate $f_{xx}(x,y),f_{yy}(x,y)$ and $f_{xy}(x,y)$ to find the nature point. Partially differentiate $\left(2\right)$ with respect to $x$

$$\begin{gathered} \frac{\partial }{\partial x}{f}_x(x,y)=\frac{\partial }{\partial x}\left\{m{y}^n{x}^{m-1}\right\} \\ {f}_{xx}(x,y)=m{y}^n\frac{\partial }{\partial x}{x}^{m-1} \\ =m(m-1)y^n{x}^{m-2} \end{gathered}$$

Partially differentiate $\left(3\right)$ with regard to $y$

$$\begin{gathered} \frac{\partial }{\partial y}{f}_y(x,y)=\frac{\partial }{\partial y}\left\{n{x}^m{y}^{n-1}\right\} \\ {f}_{yy}(x,y)=n{x}^m\frac{\partial }{\partial y}{y}^{n-1} \\ =n(n-1)x^{n\text{'}}{y}^{n-2} \end{gathered}$$

Partially differentiate $\left(2\right)$ with regard to $y$

$$\begin{gathered} \frac{\partial }{\partial y}{f}_x(x,y)=\frac{\partial }{\partial y}\left\{m{y}^n{x}^{m-1}\right\} \\ {f}_{xy}(x,y)=m{x}^{n-1}\frac{\partial }{\partial y}{y}^n \\ =mm{x}^{m-1}{y}^{n-1} \end{gathered}$$

Now, find the discriminant of $f(x, y)$

$$\begin{gathered} det\left(H_f\right)=f_{xx}{f}_{yy}-{\left(f_{xy}\right)}^2 \\ =m(m-1)y^n{x}^{m-2}·n(n-1)x^m{y}^{n-2}-{\left(mn{x}^{m-1}{y}^{n-1}\right)}^2 \\ =m(m-1)n(n-1)y^{n+n-2}{x}^{m-2+m}-m^2{n}^2{x}^{2m-2}{y}^{2n-2} \\ =mn(m-1)(n-1)y^{2n-2}{x}^{2m-2}-m^2{n}^2{x}^{2m-2}{y}^{2n-2} \\ =mn\left\{\right(m-1\left)\right(n-1)-mn\}{x}^{2m-2}{y}^{2n-2} \\ =mn(mn-m-n+1-mn)x^{2m-2}{y}^{2n-2} \\ =mn\{1-(m+n\left)\right\}{x}^{2m-2}{y}^{2n-2} \\ =mn\{1-(m+n\left)\right\}{\left(x^{m-1}{y}^{n-1}\right)}^2 \end{gathered}$$

Step 4: Calculation

The function $f(x, y)$ will have relative minimum if $det\left(H_f\right)>0$ with $f_{xx}(x,y)>0$ or $f_{yy}(x,y)>0.$

Thus, for relative minimum

$$\begin{gathered} det\left(H_f\right)>0 \\ mn\{1-(m+n\left)\right\}{\left(x^{m-1}{y}^{n-1}\right)}^2>0 \end{gathered}$$

Since $m$ and $n$ are both positive, so $mn$ is positive. Also ${\left(x^{m-1}{y}^{n-1}\right)}^2$ is positive being square. Since $mn\{1-(m+n\left)\right\}{\left(x^{m-1}{y}^{n-1}\right)}^2>0$ so $1-(m+n)$ must be positive.

Thus,

$$\begin{gathered} 1-(m+n)>0 \\ m+n<1 \end{gathered}$$

This implies, $m<1$ and $n<1$

The relative minimum conditions are $m<1$ and $n<1$

The function $f(x, y)$ will have a saddle point if $det\left(H_f\right)<0$

Thus, for saddle point

$$\begin{gathered} det\left(H_f\right)<0 \\ mn\{1-(m+n\left)\right\}{\left(x^{m-1}{y}^{n-1}\right)}^2<0 \end{gathered}$$

Since $m$ and $n$ are both positive, so $mn$ is positive. Also ${\left(x^{m-1}{y}^{n-1}\right)}^2$ is positive being square. Since $mn\{1-(m+n\left)\right\}{\left(x^{m-1}{y}^{n-1}\right)}^2<0$, so $1-(m+n)$ must be negative.

Thus,

$$\begin{gathered} 1-(m+n)<0 \\ m+n>1 \end{gathered}$$

The condition for saddle point is $m+n>1$