Ex.1

We can extend the stationary condition to partial derivatives,
and we have \(\frac{\partial f}{\partial x}\) = 0 and \(\frac{\partial f}{\partial y}\) = 0.
In this case, we have 2x + y = 0 and 2y + x = 0. The first condition gives y = -2x and the second condition gives x = -2y.
Substituting -2x = y to the second condition, which we get x = 4x. The values satisfies the condition is 0.
Now we substitute x = -2y to the first condition, which we get y = 4y. The value satisfy the condition is y = 0.
Now how do we know that it corresponds to a maximum or minimum?
If we try to use the second derivatives, we have four different partial derivatives and need to define Hessian matrix from these second partial derivatives, and we have the following:
\(\ H = 2\times2 - 1\times1 = 3\) is a positive definite, then the stationary point (0, 0) corresponds to a local minimum. So \(f(x)_{min} = 0\).

Ex. 2

We can see from the below curve that the y values is increasing as x value is increasing.
So the new minimum is 1 when the new constraint \(x^{2} \ge 1\). \(f(x)_{min} = 1\)

Ex. 3

From \(f = x^2+2xy+y^2 and\ y = x^2-2\), we have
\(\phi = f + \lambda h = x^2 +2xy + y^2 + \lambda(x^2 - y - 2)\)
So we have
\(\displaystyle \frac{\partial \phi}{\partial x} = x + y + \lambda x = 0\)
\(\displaystyle \frac{\partial \phi}{\partial y} = 2x + 2y - \lambda = 0\)
\(\displaystyle \frac{\partial \phi}{\partial \lambda} = x^2 - y - 2 = 0\)
The first condition gives \(x+y = - \lambda x\). which substitute the second condition.
That gives us \(-2\lambda x= \lambda\)
means either \(\lambda = 0 \ or\ x = -\frac{1}{2}\)
If \(x = -\frac{1}{2}\), then \(y = -\frac{7}{4}\) will satisfy all three conditions. Then \(f(-\frac{1}{2}, -\frac{7}{4}) = 5.0625\)
If \(\lambda = 0\), then \(y^2 - y -2 = 0\), which gives us \(x = -2, y=2\) or \(x=1, y=-1\)
Substitute \((-2,2)\) to \(f = x^2+2xy+y^2\), we get \((-2)^2 + 2(-2)(2) +2^2 = 0\)
Substitute \((1,-1)\) to \(f = x^2+2xy+y^2\), we get \((1)^2 + 2(1)(-1) +(-1)^2 = 0\)
So we are having \(f(x)_{min} = 0\)