Find the minimum of \(f(x,y) = x^{2} + xy + y^{2} in (x,y) \in R^{2}\)
Solution:
The stationary conditions are \(\frac{\partial f}{\partial x} =2x + y=0\) and \(\frac{\partial f}{\partial y} =x + 2y=0\)
From the second condition we get either x = -2y or y = \(\frac{-x}{2}\)
Substituting x = -2y into the first condition gives us
2(-2y) + y = 0
-4y + y = 0
-3y = 0
y = 0
Substituting y = \(\frac{-x}{2}\) into the first condition gives us
2x + \(\frac{-x}{2}\) = 0
\(\frac{3x}{2}\) = 0
x = 0
So x = 0 and y = 0. These also hold true for the second condition.
Hessian Matrix \(H = \begin{pmatrix} f_{xx} & f_{xy} \\f_{yx} & f_{yy} \end{pmatrix}\) = \(\begin{pmatrix} 2 & 1 \\1& 2 \end{pmatrix}\)
fxx = 2x + y = 2
fxy = 2x + y = 1
fyx = x + 2y = 1
fyy = x + 2y = 2
H is positive definite which means that the point (0,0) is a minimum.
For \(f(x)= x^{4}inR,\) it has a global minimum at x = 0. Find its new minimum if a constraint \(x^{2} \geq 1\) is added.
Solution:
\(f(x)= x^{4}\) and \(g(x)= x^{2}\)
\(\amalg = f(x) + \mu [g(x)]^2\) = \(x^{4} + \mu [x^{2}]^2\) = \(x^{4} + \mu [x^{4}]\)
\(\amalg'(x) = 4 x^{3} + 4 \mu x^{3} = 0\) = \(\amalg'(x) = 4 x^{3} (1 + \mu) = 0\)
X can be either 1 or -1. The new minimum is x = 1.
Use a Lagrange multiplier to solve the optimization problem \(min f(x,y) = x^{2} + 2xy + y^{2}\), subject to \(y = x^{2} - 2\)
Solution:
\(L = f(x, y) + \lambda h(x,y)\) \(h(x, y) = x^{2} - 2 - y\)
\(L = x^{2} + 2xy + y^{2} + \lambda (x^{2} - 2 - y)\)
\(\frac{\partial L}{\partial x} = 2x + 2y + 2\lambda x = 0\), \(\frac{\partial L}{\partial y} = 2x + 2y - \lambda = 0\), \(\frac{\partial L}{\partial \lambda } = x^{2} - 2 - y = 0\)
For the second condition, \(2x + 2y - \lambda = 0\) gives us \(\lambda = 2(x + y)\)
Plugging \(\lambda = 2(x + y)\) into the first condition gives us
\(2x + 2y + 2(2(x + y)) x = 0\)
\(2x + 2y + 4x^2 + 4xy = 0\)
\(2(x + y + 2x^2 + 2xy) = 0\) which gives us \(y = \frac{-2x^2 - x}{2x + 1}\)
Substitute \(y = \frac{-2x^2 - x}{2x + 1}\) into the third condition gives us
\(x^{2} - 2 - \frac{-2x^2 - x}{2x + 1} = 0\) when simplified gives x = 1 or x = - 2.
Plugging x = 1 into the third condition gives us y = -1 and plugging in x = -2 gives us y = 2. We get two points (1, -1) and (-2, 2).
The optimality for \(min f(x,y) = x^{2} + 2xy + y^{2}\) is (1, -1) with \(f_{min} = 0\) and (-2, 2) with \(f_{min} = 0\)