Answer:
The absolute maximum of f(x,y) is 121. The absolute minimum of f(x,y) is -121
Explanation:
The given function f(x,y) can be seen as a quadratic form:
![f(x,y)=(x,y)\left[\begin{array}{cc}1&0\\0&-1\end{array}\right](x,y)^(T)=X^(T)AX](https://img.qammunity.org/2021/formulas/mathematics/college/x7lgzd2l1duxf8gr5or34udh3xk7o5vjd0.png)
The constraint can be seen as:
![f(x,y)=(x,y)\left[\begin{array}{cc}1&0\\0&1\end{array}\right](x,y)^(T)=121\\X^(T)IX=X^(T)X=|X|^2=121](https://img.qammunity.org/2021/formulas/mathematics/college/guv93pcr6dj1bu1xksqb6cwc5gof3cgbz7.png)
Using the Min-max theorem with Rayleigh–Ritz quotient, we can easly obtain the absolute maximum and minimum of a quadratic form:
![\lambda_(minf)\leq (f(x,y))/(|X|^2) \leq \lambda_(Maxf)](https://img.qammunity.org/2021/formulas/mathematics/college/df29joxv7u25flcyz85eydmzkml07gs9eo.png)
Therefore:
![\left \{ {X \atop X} \right.](https://img.qammunity.org/2021/formulas/mathematics/college/khzejibz23kzipa8x6n8wwoulpfmg1gatm.png)
So the problem is reduced to obtain the maximum and minimum eigenvalues of the matrix A.
This eigenvalues can be obtained directly (diagonal matrix), where
and
. Therefore:
![\left \{ {^2=1 \cdot 121=121 \atop min(f(X)=\lambda_(minf)} \right.](https://img.qammunity.org/2021/formulas/mathematics/college/rp2beuxqv04wr9hdp9z8t2zzcc8qsha2xj.png)