prove that for all real x , y the value of
x^2 + 2xy + 3y^2 -6x-2y
cannot be less than-11
( edited )
2.
solve
( x- 3 )^5 + ( x-1 )^5\geq 244
for prctc
-
UP 0 DOWN 0 1 3
3 Answers
39For Q1,
Let z = x² + 2xy + 3y² - 6x - 2y
We will use partial differentiaton to find dy/dx.
Diff wrt x taking y as constant.
Fx = 2x + 2y + 0 - 6 - 0 = 2x + 2y - 6
Similarly diff wrt y taking x as constant,
Fy = 0 + 2x + 6y - 0 - 2 = 2x + 6y - 2
Now dy/dx = -Fx/Fy = 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!
1708alternate:-->(1)
means we have to prove
x2+2xy+3y2-6x-2y>= - 11 or
x2+2xy+3y2-6x-6y+4y +(3)2-(3)2
x2+2xy+y2+(3)2-2*3(x+y)+2y2+4y+2-2-(3)2
so
(x+y-3)2+2(y-1)2 -11>=-11
