Mathematics: Find the absolute maximum and minimum of the function f(x,y)=x^2−y^2f(x,y)=x^2−y^2 subject to the constraint x^2+y^2=121x^2+y^2=121.

slkfnvsoifnfbd;lknbdlknb;ldknbglksStep-by-step explanation:...

Find the absolute maximum and minimum of the function

kalebrayburn62

14.01.2022

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:

$f(x,y)=(x,y)\left[\begin{array}{cc}1&0\\0&-1\end{array}\right](x,y)^{T}=X^{T}AX$

The constraint can be seen as:

$f(x,y)=(x,y)\left[\begin{array}{cc}1&0\\0&1\end{array}\right](x,y)^{T}=121\\X^{T}IX=X^{T}X=|X|^2=121$

Using the Min-max theorem with Rayleigh–Ritz quotient, we can easly obtain the absolute maximum and minimum of a quadratic form:

$\lambda_{minf}\leq \frac{f(x,y)}{|X|^2} \leq \lambda_{Maxf}$

Therefore:

$\left \{ {{Max(f(X)=\lambda_{maxf}|X|^2} \atop {min(f(X)=\lambda_{minf}|X|^2}} \right.$

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 $\lambda_{minf}=-1$ and $\lambda_{Maxf}=1$ . Therefore:

$\left \{ {{Max(f(X)=\lambda_{maxf}|X|^2}=1 \cdot 121=121 \atop {min(f(X)=\lambda_{minf}|X|^2=-1 \cdot 121=-121}} \right.$

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