Find all values of x and y such that fx(x, y) = 0 and fy(x, y) = 0 simultaneously.f(x, y) = x² + 4xy + y² - 34x - 32y + 38

Solution:

Given: f(x, y) = x² + 4xy + y² - 34x - 32y + 38

By differentiating it with respect to x

f\(_x\) = 2x + 4y - 34 = 0

By differentiating it with respect to y

f\(_y\) = 4x + 2y - 32 = 0

It can be written as

2x + 4y = 34 --- (1)

4x + 2y = 32 --- (2)

Let us solve the system of linear equations

Now multiply the equation (1) by -2

-4x - 8y = - 68 --- (3)

By adding (2) and (3)

-6y = -36

Divide both sides by -6

y = 6

Substitute the value of y in equation (1)

2x + 4(6) = 34

By further calculation

2x + 24 = 34

2x = 34 - 24

2x = 10

Divide both sides by 2

x = 5

Therefore, the values of x and y are 5 and 6.

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Find all values of x and y such that fx(x, y) = 0 and fy(x, y) = 0 simultaneously.f(x, y) = x² + 4xy + y² - 34x - 32y + 38

Summary:

The values of x and y such that fx(x, y) = 0 and fy(x, y) = 0 simultaneously, f(x, y) = x² + 4xy + y² - 34x - 32y + 38 are 5 and 6.