1) \[F(x,y)=x^2+xy+y^2\]
To Find the minimum of the polynomial above, we first have take the partial derivatives of the polynomial and input into the Hessian matrix.
First, let’s find the partial derivative of x and y
\[\frac{df}{dx}=2x+y\] \[\frac{df}{dy}=x+2y\]
When we look at the minimum solution, we can show that f(0,0)=0 will be the case as x=0 and y=0. \[\frac{df}{dx}=2x+y=0\] \[\frac{df}{dy}=x+2y=0\]
Then, we will take the second partial derivative below for all terms.
\[\frac{df}{dxx}=2\] \[\frac{df}{dyy}=2\] \[\frac{df}{dxy}=1\] \[\frac{df}{dyx}=1\]
We can insert the results of the partials in the Hessian Matrix. The Hessian matrix results in a positive definite of 3, so (0,0) is the minimum \[H= \binom{21}{12}=4-1=3 \]
2) Finding new global minimum with constriant
(This method is very new to me so tracking to what is in the lecture) Following on, we have this function x^4 that will be to minimize but it has a constraint x^2. \[f(x)=x^4\]
we are aware that our constraint is larger than one so we can built out the unconstrained problem \[(x,u)=f(x)+u(g(x))=x^4+u(x^2)^2->x^4+ux^4\]
Then we take the derivative of the unconstrained to be \[p'(x)=4x^3+4ux^3=0->4x^3(1+u)=0\]
(I am stuck at this part.I don’t know how to proceed. if I solve for x, I get zero.)
3) Solve for the min of \[x^+2xy+y^2\] subject to y=x^2-2 using the langrange multiplier
For this problem, we have to transform the given constraint into its form \[g(x,y)=C->y=x^2-2 ->x^2-2-y=0\]
Now, we can form the langrangian to be
\[L(x,y)=x^2+2xy+y^2+\lambda(x^2-2-y)\]
We can take the partial derivatives for the stationary conditions \[\frac{dL}{dx}=2x+2y-2x\lambda\] \[\frac{dL}{dy}=2x+2y-\lambda\] \[\frac{dL}{d\lambda}=x^2- 2-y\]
Using the derivative of y, we can solve for lambda to be \[2x+2y-\lambda=0->\lambda=2x+2y->\lambda=2(x+y)\]
Then, let’s apply the substitution for lambda in the derivative of x \[0=2x+2y-2x(2x+2y)->2x+2y+4x^2+4xy=0->2(x+y+2x^2+2xy)=0->x+y+2x^2+2xy=0->y+2xy=-2x^2-x->y(1+2x)=-2x^2-x->y=\frac{-2x^2-x}{1+2x}\]
Now, we can find out x using the derivative for \[0=x^2- 2-\frac{-2x^2-x}{1+2x}->x^2-2+x=0-> (x-1)(x+2)=0\]
we get the x=-2,1 from solving. When we plug into our last derivative \[0=(-2)^2- 2-y->y=2|0=(1)^2-2-y->y=-1\] so, we receieve the most optimize points of (-2,2) and (1,-1)