91Ó°ÊÓ

A function is given as $z=f(x,y)=(y-x^2)(y-2x^2)$.

To evaluate maximum and minimum points of a function $\mathit{f}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}$the following points should be noted:

1. ${\mathit{f}}_{\mathbf{x}}\mathbf{=}\mathbf{0}\mathbf{=}{\mathit{f}}_{\mathbf{y}}$At $\mathbf{(}\mathit{a}\mathbf{,}\mathit{b}\mathbf{)}$.

2. role="math" localid="1659153407539" $\mathbf{(}\mathit{a}\mathbf{,}\mathit{b}\mathbf{)}$is a minimum point if ${\mathit{f}}_{\mathbf{x}\mathbf{x}}\mathbf{>}\mathbf{0}\mathbf{,}\mathbf{ }{\mathit{f}}_{\mathbf{y}\mathbf{y}}\mathbf{>}\mathbf{0}\mathbf{,}$and ${\mathit{f}}_{\mathbf{x}\mathbf{x}}{\mathit{f}}_{\mathbf{y}\mathbf{y}}\mathbf{>}{{\mathit{f}}^{\mathbf{2}}}_{\mathbf{x}\mathbf{x}}$.

3. $\mathbf{(}\mathit{a}\mathbf{,}\mathit{b}\mathbf{)}$ is a maximum point if ${\mathit{f}}_{\mathbf{x}\mathbf{x}}\mathbf{<}\mathbf{0}\mathbf{,}\mathbf{ }{\mathit{f}}_{\mathbf{y}\mathbf{y}}\mathbf{<}\mathbf{0}$, and .${\mathit{f}}_{\mathbf{x}\mathbf{x}}{\mathit{f}}_{\mathbf{y}\mathbf{y}}\mathbf{>}{{\mathit{f}}^{\mathbf{2}}}_{\mathbf{x}\mathbf{y}}$

4. $\mathbf{(}\mathit{a}\mathbf{,}\mathit{b}\mathbf{)}$is neither a maximum nor a minimum point if ${\mathit{f}}_{\mathbf{xx}}{\mathit{f}}_{\mathbf{yy}}\mathbf{>}{{\mathit{f}}^{\mathbf{2}}}_{\mathbf{xy}}$ that is ${\mathit{f}}_{\mathbf{yy}}$and ${\mathit{f}}_{\mathbf{xx}}$are of opposite sign.

Here,$f_x , f_y$are the first order partial derivatives and $f_{xx} , f_{yy}, f_{xy}$are the second order partial derivatives.

Consider the given function as follows:

$z=(y-x^2)(y-2x^2)$

$f(x,y)=y^2-3x^{2}y+2x^4$ …â¶Ä¦.(1)

Differentiate (1) partially with respect to x as shown below.

role="math" localid="1659153814205" $f_x(x,y)=\frac{∂}{∂x}(y^2-3x^{2}y+2x^4)$

$f_x(x,y)=\frac{∂}{∂x}{y}^2-\frac{∂}{∂x}3x^{2}y+\frac{∂}{∂x}2x^4$

$f_x(x,y)=-6xy+8x^3$ ……. (2)

Differentiate (1) partially with respect to y as shown below.

$f_y(x,y)=\frac{∂}{∂y}(y^2-3x^{2}y+2x^4)$

$f_y(x,y)=\frac{∂}{∂y}{y}^2-\frac{∂}{∂y}3x^{2}y+\frac{∂}{∂y}2x^4$

$f_y(x,y)=2y-3x^2$ …â¶Ä¦.(3)

Equate equation (2) with zero as follows:

$$\begin{gathered} f_x\left(x,y\right)=-6xy+8x^3 \\ 0=-6xy+8x^3 \end{gathered}$$ …â¶Ä¦.(4)

Now, equate equation (3) with zero as :

$$\begin{gathered} f_x\left(x,y\right)=2y-3x^2 \\ 0=2y-3x^2 \\ 2y=3x^2 \end{gathered}$$

$y=\frac{3x^2}{2}$ …â¶Ä¦.(5)

Solve equation (4) and (5) to obtain the value of x.

$$\begin{gathered} 8x^3=6xy \\ 6x\left(\frac{3x^2}{2}\right)=8x^3 \\ 9x^3=8x^3 \\ x=0 \end{gathered}$$

Now, solve to obtain the value of y.

$$\begin{gathered} y=\frac{3\left(0\right)}{2} \\ y=0 \end{gathered}$$

The obtained point is $\left(0,0\right)$.

Now, differentiate (2) and (3) again as shown below:

$f_{xx}(x,y)=-6y+24x^2$ …â¶Ä¦.(6)

$f_{yy}(x,y)=2$ …â¶Ä¦(7)

$f_{xy}(x,y)=-6x$ …â¶Ä¦.(8)

Now, use the value obtained in equation and (8) to check the maximum and minimum point.

At $\left(0,0\right)$,$f_{xx}=0 , f_{yy}>0$and $f_{xx}{f}_{yy}={f_{xy}}^2$.

Therefore, the point $\left(0,0\right)$is neither minimum nor maximum point of the given function.

Now, consider any straight line through origin as$y=mx$.

Substitute the above in given function as follows:

$$\begin{gathered} z=f\left(x,mx\right) \\ z=\left(mx-x^2\right)\left(mx-2x^2\right) \\ z=2x^4-3m{x}^3+m^2{x}^2 \end{gathered}$$

Differentiate partially with respect to xas follows:

$$\begin{gathered} \frac{∂z}{∂x}=f_x\left(x,mx\right) \\ \frac{∂z}{∂x}=\frac{∂}{∂x}\left(2x^2-3m{x}^3+\frac{∂}{∂x}{m}^2{x}^2\right) \\ \frac{∂z}{∂x}=8x^3-9m{x}^2+2m^{2}x \end{gathered}$$

Differentiate partially with respect tox as shown below.

$$\begin{gathered} \frac{{∂}^{2}z}{∂x^2}=f_{xx}(x,mx) \\ \frac{{∂}^{2}z}{∂x^2}=\frac{∂}{∂x}(8x^3-9m{x}^2+2m^{2}x) \\ \frac{{∂}^{2}z}{∂x^2}=\frac{∂}{∂x}8x^3-\frac{∂}{∂x}9m{x}^2\frac{∂}{∂x}2m^{2}x \\ \frac{{∂}^{2}z}{∂x^2}=24x^2-18mx+2m^2 \end{gathered}$$

Therefore, at $\left(0,0\right)$,$\frac{∂z}{∂x}=0$and$\frac{{∂}^{2}z}{∂x^2}=2m^2>0$.

So, at $\left(0,0\right)$the function z has minimum through a straight line.