Find the minimum of \(f(x,y)=x^2+xy+y^2\)in\((x,y) \in R^2\)
Two first partial derivatives \(\displaystyle \frac{\partial f}{\partial x}\) and \(\displaystyle \frac{\partial f}{\partial y}\) thus the stationary conditions means: \(\displaystyle \frac{\partial f}{\partial x} = 2x +y = 0\), \(\displaystyle \frac{\partial f}{\partial y} = x + 2y = 0\)
Which can be written compactly as a gradient vector \(\Delta f = (\displaystyle \frac{\partial f}{\partial x}, \displaystyle \frac{\partial f}{\partial y})=0\)
from the second condition we get either, \(x = -2y\) or \(y = \frac{-x}{2}\) substituting \(x=-2y\) to first condition gives us,
\(2(-2y)+y=0\)
\(-4y+y=0\)
\(-3y=0\)
\(y=0\)
substituting \(y = \frac{-x}{2}\) to first condition gives us,
\(2x+\frac{-x}{2}=0\)
\(\frac{3x}{2}=0\)
\(x=0\)
Therefor \(x=0, y=0\)
We can get the second partial derivatives to form Hessian Matrix
\(f_{xx} = \displaystyle \frac{\partial^2 f}{\partial x^2}, f_{yy}=\displaystyle \frac{\partial^2 f}{\partial y^2}, f_{xy}=\displaystyle \frac{\partial^2 f}{\partial x\partial y}, f_{yx}=\displaystyle \frac{\partial^2 f}{\partial y\partial x}\)
\(f_{xx}=2x+y=2\)
\(f_{xy}=2x+y=1\)
\(f_{yx}=x+2y=1\)
\(f_{yy}=x+2y=2\)
\[H = \left(\begin{array}{cc} 2 & 0\\ 0 & 2 \end{array}\right) \] We get the determinant of Hessian Matrix which is (4-2=2) and it is positive definite. In that case point \((0,0)\) corresponds to a minimum with \(f_{min} = 0\).
For \(f(x) = x^4 \in R\), it has the global minimum at \(x=0\). Find its new minimum if a constraint \(x^2 \ge 1\) is added.
\(f(x) = x^4\)
\(g(x) = x^2\)
\(\amalg = f(x) + \mu[g(x)]^2\) = \(x^4 + \mu[x^2]^2\) = \(x^4 + \mu[x^4]\)
\(\amalg` = 4x^3 + 4\mu x^3 = 0\) = \(4x^3 (1+\mu) = 0\)
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 \(y=x^2-2\)
\[L = f(x,y)+\lambda h(x,y)\]
\(h(x,y) = x^2-2-y\)
\(L = x^2+2xy+y^2+\lambda(x^2-2-y)\)
\(\frac{\partial L}{\partial x}=2x+2y+2\lambda x =0, \frac{\partial L}{\partial y}=2x+2y-\lambda =0, \frac{\partial L}{\partial \lambda}=x^2-2-y=0\)
For the second condition, \(2x+2y-\lambda=0\) gives us \(\lambda = 2(x+y)\) We can plug \(\lambda = 2(x+y)\) to our first condition:
\(2x+2y+2(2x+2y)x =0\) \(2x+2y+(4x+4y)x\) \(2x+2y+4x^2+4xy=0\)
\(2(x+y+2x^2+2xy)=0\) gives us,
\(y = \frac{-2x^2-x}{2x+1}\)
substituting y into third condition gives us,
\(x^2-2-\frac{-2x^2-x}{2x+1}=0\)
Once you simplify you get
\(x^2-2+x=0\) where \(x =1\) and \(x=-2\)
Plugging both x in third condition gives us: \(x=1, 1^2-2-y=0\) gives us \(y=-1\)
\(x=-2, -2^2-2-y=0\) gives us\(y=2\)
two points we get are (1,-1) & (-2,2).
Therefore, optimality occurs at (1,-1) and (-2,2) with \(f_{min}=0\)