1) Find the minimum of \(f(x,y)=x^2 +xy +y^2 \space in \space (x,y) \in \mathbb {R}^2\)

\(f_x= 2x +y\)
\(f_y= x + 2y\)
\(f_{xx} = 2\)
\(f_{xy} = 1\)
\(f_{yy} = 2\)
\(x + 2y = 0\)
\(2x + y = 0\) \(x = -2y\)
\(2(-2y) + y = 0\)
\(-4y + y = 0\)
\(-3y = 0\)
\(y=0 x=0\)

We have a stationary point at (0,0)

\(f_{xx}f_{yy}-f^2_{xy}\) = \((2)(2)-(1)\) = 3 is greater than 0 so (0,0) is either a minimum or maximum.

\(f_{xx}(0,0) = 2 > 0\)

\(f_{yy}(0,0) = 2 > 0\)

Therefore, (0,0) is the minimum point to the function \(f(x,y)=x^2 +xy +y^2 \space in \space (x,y) \in \mathbb {R}^2\)

           

2) For \(f(x)=x^4 \space in \space (x,y) \in \mathbb {R}\), it has the global minimum at x = 0. Find its new minimum if a constraint \(x^2 \ge 1\) is added.

\(f'(x) = 4x^3\) has a critical point at (0,0) and is decreasing from \((-\infty,0)\) and increasing from \((0,\infty)\).

The minimum should then occur at the beginning of the constraint which is at x = 1 or x = -1.

The values of which are at (1,1) and (-1,1).

         

3) Use a Lagrange multiplier to solve the optimization problem min \(f(x,y)=x^2+2xy+y^2\) subject to \(y=x^2 - 2\)

    \(\frac{df}{dx}=\lambda \frac{dg}{dx}\)

\(\frac{df}{dy}=\lambda \frac{dg}{dx}\)

\(2x + 2y = 2 \lambda x\)
\(2x = -2y\)
\(x = -y\)
\(y=x^2 - 2\)
\(\lambda = \frac{x+y}{x}\)
\(-x=x^2 - 2\)
\(x^2 +x - 2 = 0\)
\(x=-2\) or \(x=1\)
Thus we have minimums at points (2,-2) and (1,-1)