prove

For Q1,
Let z = x² + 2xy + 3y² - 6x - 2y
We will use partial differentiaton to find dy/dx.
Diff wrt x taking y as constant.
F_x = 2x + 2y + 0 - 6 - 0 = 2x + 2y - 6
Similarly diff wrt y taking x as constant,
F_y = 0 + 2x + 6y - 0 - 2 = 2x + 6y - 2

Now dy/dx = -F_x/F_y = 3 - x - yx + 3y - 1
Now this is zero when x + y = 3. So the function changes sign across (x + y) = 3.
Now at y = 3 - x the function should give its minimum or maximum values.
z = x² + 2x(3 - x) + 3(3 - x)² - 6x - 2(3 - x)
= x² + 6x - 2x² + 27 + 3x² - 18x - 6x - 6 + 2x
= 2x² - 16x + 21
Now how to proceed I don't know.
z/2 = x² - 8x + 21/2
= x² - 8x + 16 - 16 + 21/2
z/2 = (x - 4)² - 11/2
=> z/2 + 11/2 = (x - 4)²
Now as (x - 4)² ≥ 0 for all x,
z ≥ -11
SOLVED!

still one more method left !!!!!!! for first one !!