Minimum value of f(x,y)=x^2-8xy+9y^2-16y+10

[kinu]

If \(f(x,y) = x^2-8xy+9y^2-16y+10\), Where \(x,y\in \mathbb{R}\)

Then Minimum value of the expression

[josesousa]

You should first comute the gradient of the function and find the zeros.
\(\nabla f =[0 ; 0]\)

In the zero(s) (in this case one), study the hessian matrix to see if it is positive definite.

[João P. Ferreira]

Dear kinu

As Professor José Sousa said you first need to compute

\(\nabla f = [0;0]\)

which is equivalent to write

\(\begin{cases}
\frac{\partial f}{\partial x}=0 \\
\frac{\partial f}{\partial y}=0 \end{cases}\)

Let's compute them...

\(\frac{\partial f}{\partial x}=2x-8y\)

\(\frac{\partial f}{\partial y}=-8x+18y-16\)

Now we need to solve this system

\(\begin{cases}
\frac{\partial f}{\partial x}=0 \\
\frac{\partial f}{\partial y}=0 \end{cases} \ \ \Leftrightarrow \ \ \begin{cases}
2x-8y=0 \\
-8x+18y-16=0 \end{cases} \ \ \Leftrightarrow \ \ \begin{cases}
x=-\frac{32}{7} \\
y=-\frac{8}{7} \end{cases}\)

Now let's find the Hessian matrix

\(H \left [ f(x, y) \right] = \begin{bmatrix}
\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x\,\partial y} \\ \\ \frac{\partial^2 f}{\partial y\,\partial x} & \frac{\partial^2 f}{\partial y^2} \\
\end{bmatrix}\)

\(\frac{\partial^2 f}{\partial x^2}=2\)

\(\frac{\partial^2 f}{\partial y^2}=18\)

\(\frac{\partial^2 f}{\partial x\,\partial y}=\frac{\partial^2 f}{\partial y\,\partial x}=-8\)

Then

\(H(x,y)=\begin{bmatrix} 2 & -8\\ -8 & 2 \end{bmatrix}\)

Because

\(H_1>0\) and \(H_2<0\) the Hessian matrix is indefinite.

\(H_1=2\)

\(H_2=\begin{vmatrix} 2 & -8\\ -8 & 18 \end{vmatrix}=-28\)

Because the Hessian matrix is indefinite, that point is a saddle point.

So that functions has no minimuns

Take care

[josesousa]

Notice also that

\(x^2-8xy+9y^2-16y+10=\)
\(=(x-4y)^2-7y^2-16y+10\)

When \(x=4y\) and \(y\) is large, this tends to \(-\infty\)
Otherwise, when \(y=0, x\) large, then it tends to \(+\infty\)

[kinu]

Thanks Admin and Professor

But would you like to give me an Idea about Hessian Matrix

and Professor josesousa Like for \(x=4y\)

It is also True for \(x=2y\)

I did not Understand The Reason Behind That.

Thanks.

Kinu

[João P. Ferreira]

Dear kinu

Regarding the Hessian matrix I strongly recommend you to read this article
http://en.wikipedia.org/wiki/Hessian_matrix

In mathematics, the Hessian matrix (or simply the Hessian) is the square matrix of second-order partial derivatives of a function; that is, it describes the local curvature of a function of many variables. The Hessian matrix was developed in the 19th century by the German mathematician Ludwig Otto Hesse and later named after him. Hesse himself had used the term "functional determinants".

[kinu]

Thanks Admin

[josesousa]

Kinu, I just gave those 2 trajectories as simple examples. The study of the limits in the infinite will depend on the trajectories considered and innumerous examples can be given.