Question

The minimum value of
\[x^{2}+2xy+3y^{2}-6x-2y\]where\[x,y \epsilon R\] is ???????

Answer 1

can we use calculus ?

Answer 2

No

Answer 3

we are not of that level

Answer 4

ok

Answer 5

well i want to use quadratic equations
i think u are more comfortable with them

Answer 6

first let \(x^2+2xy+3y^2-6x-2y=z \) we want to find the minimum of z

Answer 7

i want to arrange the equation to appear as a quadratic equation in terms of x

Answer 8

so i have \( x^2+(2y-6)x+3y^2-2y-z=0 \)

Answer 9

in three variables?

Answer 10

no problem with that

Answer 11

x is real so the discriminant of that quadratic equation must be greater than or equal to zero

Answer 12

hence \( (2y-6)^2-4(3y^2-2y-z) \ge 0 \) simplify it u will get \( 2y^2+4y-9-t \ge 0 \)

Answer 13

sorry u will get \( 2y^2+4y-9-t \le 0 \)

Answer 14

right till here?

Answer 15

oh waht a mistake its z not t

Answer 16

yes then

Answer 17

now we must have D>=0 for f(y)=2y^2+4y-9-z it will gives z>=-11
so the minimum is -11

Answer 18

how -11?

Answer 19

see if D<0 for f(y)=2y^2+4y-9-z then all of f(y) lies on the above of x-axis and then f(y)>0 so we must have D>=0

Answer 20

D is discriminant

Answer 21

how have you solved to -11?

Answer 22

discriminant of 2y^2+4y-9-z is 16+8(z+9) so 16+8(z+9)>=0 or z>=-11

Answer 23

interesting question; i've seen like this before

Answer 24

pls explain

Answer 25

why we have taken D >=0, we have to take D<=0

Answer 26

ok see if D<=0 there is no real values for y and one of the conditions of question is to be y from real numbers

Answer 27

ok understood

Answer 28

If we would have taken the eq. in terms of x then also the ans. would have come -11??

Answer 29

yes sure

Answer 30

but i am getting z >=-28 when taken it in terms of x.

Answer 31

let me check it

Answer 32

in terms of y

Answer 33

in terms of x you have solved

Answer 34

i got z>=-11 when we put it in terms of y in the very first equation

Answer 35

so its allright

Answer 36

what eq. you got in terms of y?

Answer 37

32+3z+1>=0

Answer 38

i got 2x^2-16x-1-z<=0

Answer 39

i got 2x^2-16x-1-3z<=0

Answer 40

ok thanks