Question

If x+y=2, what is the minimum value of x^2+y^2?

Answer 1

by squaring (x+y=2)\[\huge{(x+y)^2=(2)^2}\]\[\huge{x^2+y^2+2xy=4}\]\[\huge{x^2+y^2=4-2xy}\]

Answer 2

So how do you find the minimum value?

Answer 3

Is there given any value of xy ??/

Answer 4

Nope.

Answer 5

@ganeshie8 sir plz help :)

Answer 6

this is a quadratic function problem...
x+y=2 ---> y=2-x
x^2+y^2 = x^2 + (2-x)^2
let k=x^2+y^2, so
k=x^2 + (2-x)^2
k=x^2 + 4 - 4x + x^2
k=2x^2 - 4x + 4
k would be minimum when x=-b/2a=-(-4)/2*2 = 4/4 = 1
to find the minimum value of k, just plug x=1
k=2x^2 - 4x + 4
k=2(1)^2 - 4(1) + 4
k=2-4+4
k=2

Answer 7

for minimum value the two value should be close as possible ,,so here it should be 1,1..so minimum value should be 2

Answer 8

Where did the equation x=-b/2a come from?

Answer 9

it is x-vertec of quadratic function y=ax^2+bx+c

Answer 10

Ok, thank you.