Question

Find whether the function is maxima or minima at the stationary point. Z = f(x, y) = x²+ 2x y²+ 2y²​

Answer 1

Answer:

you solve it yourself because it's your question

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f = {x}^{2} + {2xy}^{2} + {2y}^{2} - - - (1)$

Differentiate partially w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{\partial f}{\partial x} = 2x + {2y}^{2}$

$\rm :\longmapsto\:\dfrac{\partial f}{\partial y} = 4xy + {4y}$

$\rm :\longmapsto\:A = \dfrac{ {\partial }^{2} f}{ {\partial x}^{2} } = 2$

$\rm :\longmapsto\:B = \dfrac{ {\partial }^{2} f}{ {\partial x}\partial y } = 4y$

$\rm :\longmapsto\:C = \dfrac{ {\partial }^{2} f}{ {\partial y}^{2} } = 4x + 4$

For stationary points,

$\rm :\longmapsto\:\dfrac{\partial f}{\partial x} = 0 \: \: and \: \: \dfrac{\partial f}{\partial y} = 0$

$\rm :\longmapsto\:2x + {2y}^{2} = 0 \: \: and \: \: 4xy + 4y = 0$

$\rm :\longmapsto\:x + {y}^{2} = 0 \: \: and \: \: 4y (x+ 1)= 0$

$\rm :\implies\:y = 0 \: \: \: or \: \: \: x = - 1$

Now, when y = 0

$\rm :\longmapsto\: {y}^{2} = 0$

$\rm :\implies\:y = 0$

Now, when x = - 1

$\rm :\longmapsto\: - 1 + {y}^{2} = 0$

$\rm :\longmapsto\: {y}^{2} = 1$

$\rm :\implies\:y = \: \pm \: 1$

So, stationary points are

$\rm :\longmapsto\:(0,0), \: ( - 1,1), \: ( - 1, - 1)$

Consider,

$\rm :\longmapsto\:AC - {B}^{2}$

$\rm  \:  =  \: \:2(4x + 4) - {(4y)}^{2}$

$\rm  \:  =  \: \:8x + 8 - 16 {y}^{2}$

Now,

Concept of Maxima and Minima

$\rm :\longmapsto\:If \: AC - {B}^{2} > 0, \: A > 0 \: then \: f \: is \: minimum$

$\rm :\longmapsto\:If \: AC - {B}^{2} > 0, \: A < 0 \: then \: f \: is \: maximum$

$\rm :\longmapsto\:If \: AC - {B}^{2} < 0, \: \: then \: point \: is \: saddle \: point.$

Consider,

$\rm :\longmapsto\:Case - 1 \: Point \: is \: (0,0)$

So,

$\rm :\longmapsto\:A = \dfrac{ {\partial }^{2} f}{ {\partial x}^{2} } = 2$

$\rm :\longmapsto\:B = \dfrac{ {\partial }^{2} f}{ {\partial x}\partial y } = 4y = 0$

$\rm :\longmapsto\:C = \dfrac{ {\partial }^{2} f}{ {\partial y}^{2} } = 4x + 4 = 4$

So,

$\rm :\longmapsto\:AC - {B}^{2} = 8 - 0 = 8$

$\rm :\longmapsto\:AC - {B}^{2} > 0 \: and \: A > 0$

$\rm :\implies\:f \: is \: minimum \: at \: (0,0)$

and

$\rm :\implies\:Minimum \: value \: is \: f(0,0) = 0$

$\red{\rm :\longmapsto\:Case - 2 \: Point \: is \: ( - 1,1)}$

$\rm :\longmapsto\:A = \dfrac{ {\partial }^{2} f}{ {\partial x}^{2} } = 2$

$\rm :\longmapsto\:B = \dfrac{ {\partial }^{2} f}{ {\partial x}\partial y } = 4y = 4$

$\rm :\longmapsto\:C = \dfrac{ {\partial }^{2} f}{ {\partial y}^{2} } = 4x + 4 = 0$

So,

$\rm :\longmapsto\:AC - {B}^{2} = 0 - 16 = - 16$

$\rm :\longmapsto\:AC - {B}^{2} < 0$

$\bf\implies \:( - 1,1) \: is \: saddle \: point.$

$\red{\rm :\longmapsto\:Case - 2 \: Point \: is \: ( - 1, - 1)}$

$\rm :\longmapsto\:A = 2$

$\rm :\longmapsto\:B = - 4$

$\rm :\longmapsto\:C = 0$

So,

$\rm :\longmapsto\:AC - {B}^{2} = 0 - 16 = - 16$

$\rm :\longmapsto\:AC - {B}^{2} < 0$

$\bf\implies \:( - 1, - 1) \: is \: saddle \: point.$