Question

if x and y are real numbers then find the minimum value of 4(x-2)^2+(y-3)^2-2(x-3)^2​

Answer 1

Answer:

Let f(x,y)=x2+4xy+6y2−4y+4, then ∇f=(2x+4y,4x+12y−4). An extreme value occurs when ∇f=0.

Hence let

2x+4y=04x+12y=4

Multiply the first equation by 2 and subtract from the second. This gives: y=−1 and x=2. f(2,−1)=4−8+6−4+4=2. The Hessian is

(24412)

with eigenvalues 7±41−−√>0, so the Hessian is positive definite and the value of 2 is a minimum.