Question

degree of homogenous function f(x,y)=√x+√y/x+y​

Answer 1

Answer:

the degree of homogeneous function is 1

Answer 2

Degree of the homogenous function = - 1/2

Given :

The function

$\displaystyle \sf{f(x,y) = \frac{ \sqrt{x} + \sqrt{y} }{x + y} }$

To find :

The degree of homogenous the function

Concept :

A function f(x,y) is said to be a homogenous function of degree n if

f(tx,ty) = tⁿ f(x,y) where t > 0

Solution :

Step 1 of 2 :

Write down the given function

Here the given function is

$\displaystyle \sf{f(x,y) = \frac{ \sqrt{x} + \sqrt{y} }{x + y} }$

Step 2 of 2 :

Find degree of homogenous the function

Take t > 0

Now

$\displaystyle \sf{f(tx,ty) }$

$\displaystyle \sf{ = \frac{ \sqrt{tx} + \sqrt{ty} }{tx + ty} }$

$\displaystyle \sf{ = \frac{ \sqrt{t} ( \sqrt{x} + \sqrt{y}) }{t(x + y)} }$

$\displaystyle \sf{ = \frac{ {t}^{ \frac{1}{2} } ( \sqrt{x} + \sqrt{y}) }{t(x + y)} }$

$\displaystyle \sf{ = \frac{ {t}^{ \frac{1}{2} - 1} ( \sqrt{x} + \sqrt{y}) }{(x + y)} }$

$\displaystyle \sf{ = \frac{ {t}^{ - \frac{1}{2} } ( \sqrt{x} + \sqrt{y}) }{(x + y)} }$

$\displaystyle \sf{ = {t}^{ - \frac{1}{2}} \frac{ ( \sqrt{x} + \sqrt{y}) }{(x + y)} }$

$\displaystyle \sf{ = {t}^{ - \frac{1}{2} } f(x,y) }$

$\displaystyle \sf{ \therefore \: \: f(tx,ty) = {t}^{ - \frac{1}{2} } f(x,y) }$

So degree of the homogenous function = - 1/2