# Find unconditional extremum of the function `z= -2x^2+18x-2xy-y^2+12y` .1) create necessary conditions.2) to find the critical points of the function.3) to determine the type of extremum at...

Find unconditional extremum of the function `z= -2x^2+18x-2xy-y^2+12y` .

1) create necessary conditions.

2) to find the critical points of the function.

3) to determine the type of extremum at critical point

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To find extremum we need to solve the following system of equations:

`(del z)/(del x)=0`

`(del z)/(del x)=0`

After we derivate the function we get

`-4x+18-2y=0`

`-2x-2y+12=0`

**Solution of this system is** `(x,y)=(3,3)` **which is our critical point.**

To find wether this point is minimum or maximum we need to determine if Hessian `H` is positive-definite or negative-definite.

`H=[[(del^2z)/(del x^2),(del^2z)/(del x del y)],[(del^2z)/(del y del x), (del^2z)/(del y^2)]]`

Since `z` is continous it follows that `(del^2z)/(del x del y)=(del^2z)/(del y del x)`.

`(del^2z)/(del x^2)=-4`

`(del^2z)/(del x del y)=-2`

`(del^2z)/(del y^2)=-2`

`H=[[-4,-2],[-2,-2]]`

`H` is negative-definite if `det(H)>0` and `a_(11)<0` and positive-definite if `det(H)>0` and `a_(11)>0`.

`det(H)=(-4)(-2)-(-2)(-2)=8-4=4>0` and `a_(11)=-4<0`

So Hessian is negative-definite, hence **point (3,3) is maximum**.

Since our function is actually a quadratic form it is easy to see that this is global maximum.