If `m` is the minimum value of `f(x , y)=x^2-4x+y^2+6y` when `x and y` are subjected to the restrictions `0lt=xlt=1 and 0lt=ylt=1,` then the value of `|m|` is________

Answer

3

Explanation

we have `f(x,y)=x^(2)+y^(2)-4x+6y`
Let `(x,y)=cos , theta sin theta)`.Then `theta in [0,pi//2]`and
`f(x,y)=f(theta)=cos^(2)theta+sin^(2)theta-4cos theta +6 sin`
`f(theta)= cos theta +4 sin theta gtforall0foralltheta in [0,pi//2]`
Therefore f`(theta)` is strictly increasing in `[0,pi//2]` thus
`f(theta)+_(min)=f(0)=1-4+0=-3`