Answer to Question #199535 in Calculus for Raj kumar

Question

Locate and classify the stationary points of the following:

(i) f(x,y) = 4xy + x⁴ - y⁴

(ii) f(x,y)= xy + 2/x + 4/y, x>0, y>0

Answer 1

Ans:-

$(i ) \ f(x,y) = 4xy + x^4 - y^4$

Partial derivative of $\ f(x,y)$ with respect to $x$

$\dfrac{\delta f}{\delta x}= 4y+4x^3$

Partial derivative of $\ f(x,y)$ with respect to $y$

$\dfrac{\delta f}{\delta y}= 4x-4y^3$

For stationary points put $\dfrac{\delta f}{\delta x}=0$ and $\dfrac{\delta f}{\delta y}=0$

from first equation $y=-x^3 \ \ and \ \$ from second expression $x=y^3$ Hence these condition are fulfilled by only one condition $(0,0)$ and this point will be The Origin .

For classified the nature of stationary point

$D =\dfrac{∂^2f}{∂x^2}\times \dfrac {∂^2f}{∂y^2}−(\dfrac{∂^2f}{∂x∂y})^2$

For the stationary point $(0,0)$

$D=-16\ \ <0$

the stationary point is a saddle point.

$(ii) \ f(x,y)= xy + \dfrac{2}{x}+ \dfrac{4}{y}$

Partial derivative of $\ f(x,y)$ with respect to $x$

$\dfrac{\delta f}{\delta x}=y-\dfrac{2}{x^2}$

Partial derivative of $\ f(x,y)$ with respect to $y$

$\dfrac{\delta f}{\delta y}=x-\dfrac{4}{y^2}$

For stationary points put $\dfrac{\delta f}{\delta x}=0$ and $\dfrac{\delta f}{\delta y}=0$

From first equation $yx^2=2$ and $xy^2 = 4$ Hence these condition are fulfilled by only one condition and this point will be $(1,2)$

For classified the nature of stationary point

$D =\dfrac{∂^2f}{∂x^2}\times \dfrac {∂^2f}{∂y^2}−(\dfrac{∂^2f}{∂x∂y})^2$

For the stationary point $(1,2)$

$\dfrac{∂^2f}{∂x^2}=\dfrac{4}{x^3} \ \ , \ \ \dfrac{∂^2f}{∂y^2}=\dfrac{8}{y^3} \ \ and\ \ \dfrac{∂^2f}{∂x∂y}=1$

$D=3>0 \ \ \ and \ \ \dfrac{∂^2f}{∂x^2}>0$

Then the stationary point is a local minimum.