Question

Find the STATIONARY POINTS of f(x, y) = x²– xy+y²-2x+y​

Answer 1

3xy +6

Step-by-step explanation:

x,y =x²-xy +y²-2x+y X y is equal to x square where y is equal to X when you subtract m by the magnification rule 2 X + y 2 X + 1 and your final value will be 3xy+6. thank you hope this answer helps you ,please Mark as brainliest answer.

Answer 2

The stationary point of $\bf f(x, y) = {x}^{2} – xy+ {y}^{2} -2x+y$ is (1,0).

Given:

• $f(x, y) = {x}^{2} – xy+ {y}^{2} -2x+y$

To find:

• Find the stationary points of function.

Solution:

Concept / formula to be used:

• Find f'(x) and f'(y).
• Equate both to zero.
• Solve both equations for x and y.

Step 1:

Differentiate f(x,y) with respect to x.

$\bf f'(x) = 2x - y - 2...eq1$

Differentiate f(x,y) with respect to y.

$\bf f'(y) = - x + 2y + 1...eq2$

Step 2:

Equate both eqs 1 and 2 with zero.

$f'(x) = 0$

So,

$2x - y - 2 = 0$

or

$\bf 2x - y = 2...eq3$

and

$f'(y) = 0$

So,

$- x + 2y + 1 = 0$

or

$\bf x - 2y = 1...eq4$

Step 3:

Solve eqs 3 and 4.

Multiply eq3 by 2 and subtract both

$4x - 2y = 4 \\ x - 2y = 1 \\ ( - ) \: ( + ) \: ( - ) \\ - - - - - - - \\ 3x = 3 \\ - - - - - - -$

or

$\bf \red{x = 1}$

put value of x in eq3

$2(1) - y = 2$

or

$- y = 2 - 2$

or

$\bf \red{y = 0}$

Thus,

Stationary point of f(x,y) is (1,0).

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