Answer to Question #176378 in Calculus for Sarita bartwal

Question

Check whether the function: f: R^2→R, defined by

f(x,y)= x+ysinx has extremum at any point in the domain of f.

Answer 1

$\bigstar$

The function is defined in domain

$\boxed{f(x,y)= x+ysinx}$ .......(1)

$\bigstar$ solution

=

$\frac{\partial x+ysin(x)}{\partial x}=0\\ \frac{\partial (x+ysin(x))}{ \partial y}=0$ ..................(2)

In equation (1) we do partial differentiation

w.r.t both x and y

w.r.t x we get

1 + ycos(x)= 0...............(3)

And

w.r.t y

0 + sin(x) = 0..........(4)

And using different equations (1),(2),(3),(4)

we got

$\boxed{x = 2n\pi}$ and $\boxed{y= -1}$ and $n\in z$

$\boxed{x = 2n\pi + \pi}$ and $\boxed{y= 1}$ and n$\in$ z

On partial differentiating w.r.t x and y

we got above values.

for better understand

we have different graphs

3D graph

For the function

$\boxed{f(x,y)= x+ysinx}$

CONTOUR PLOT for the function

$\boxed{f(x,y)= x+ysinx}$