Answer:
The absolute maximum of f(x,y) is 121. The absolute minimum of f(x,y) is -121
Step-by-step explanation:
The given function f(x,y) can be seen as a quadratic form:
[tex]f(x,y)=(x,y)\left[\begin{array}{cc}1&0\\0&-1\end{array}\right](x,y)^{T}=X^{T}AX[/tex]
The constraint can be seen as:
[tex]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[/tex]
Using the Min-max theorem with Rayleigh–Ritz quotient, we can easly obtain the absolute maximum and minimum of a quadratic form:
[tex]\lambda_{minf}\leq \frac{f(x,y)}{|X|^2} \leq \lambda_{Maxf}[/tex]
Therefore:
[tex]\left \{ {{Max(f(X)=\lambda_{maxf}|X|^2} \atop {min(f(X)=\lambda_{minf}|X|^2}} \right.[/tex]
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 [tex]\lambda_{minf}=-1[/tex] and [tex]\lambda_{Maxf}=1[/tex]. Therefore:
[tex]\left \{ {{Max(f(X)=\lambda_{maxf}|X|^2}=1 \cdot 121=121 \atop {min(f(X)=\lambda_{minf}|X|^2=-1 \cdot 121=-121}} \right.[/tex]
