1. algebra probs

Problem:
$f(x,) = x^2+x(y^2-1)$

a) find the extremes and classify them.
b) now, $x^2+x^2 \leq 17$ Calculate the highest and lowest f(x,y)

Skärmavbild 2012-09-11 kl. 19.42.48.png

Attempt:
a)
2012-09-11 19.39.20.jpg
b)
2012-09-11 19.39.50.jpg

The thing is this: the anser is that x = -√17 is also an alternative. But I got x = 8/3 och x = -2.

2. Re: algebra probs

Originally Posted by jacob93
Problem:
$f(x,) = x^2+x(y^2-1)$

a) find the extremes and classify them.
b) now, $x^2+x^2 \leq 17$ Calculate the highest and lowest f(x,y)

Skärmavbild 2012-09-11 kl. 19.42.48.png

Attempt:
a)
2012-09-11 19.39.20.jpg
b)
2012-09-11 19.39.50.jpg

The thing is this: the anser is that x = -√17 is also an alternative. But I got x = 8/3 och x = -2.
We can use the second derivative test to find the extreem values.

$\frac{\partial f}{\partial x}=2x+(y^2-1)=0$

and

$\frac{\partial f}{\partial y}=2xy=0$

The 2nd equation gives rise to two different cases x=0 and y=0.

Case I: if x=0 then

$y^2-1=0 \implies y=\pm1$

Case II: If y=0 then,

$2x-1=0 \implies x=\frac{1}{2}$

Now we can make the Hessian matrix

$H(x,y)=\begin{vmatrix}f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix}=\begin{vmatrix}2 & 2y \\ 2y & 2x \end{vmatrix}=4x-4y^2$

Now we can check the critical points found above

Case I: $(0,\pm 1)$

$H(0, \pm 1)=4(0)-4(\pm 1)^2=-4$

So both of these are Saddle points.

Case I: $\left(\frac{1}{2}, 0 \right)$

$H\left( \frac{1}{2},0 \right)=4\cdot \frac{1}{2}-4(0)^2=2$

Since both eigen values of the Hessian are positive the point in a local minimum.

For Part b) all we need to do is calculate the maximum and minimum on the boundary.

Since the boundary is a circle with radius $\sqrt{17}$, it can be parameterized by

$x=\sqrt{17}\cos(t) \quad y=\sqrt{17}\sin(t)$

If we put these into the original function we get

$f(t) = 17 \cos^2(t)-\sqrt{17}\cos(t)(17\sin^2(t)-1)$

This is a function of one variable and its extreem values can be found using methods from first semester calculus.

Work not shown, but you will find that $t=\pi n, \quad n \in \mathbb{Z}$ is a critical point. If you put this back into the parameteric equations you get

$x=\sqrt{17}\cos(\pi n)=\pm \sqrt{17} \quad y=\sqrt{17}\sin(\pi n)=0$.

3. Re: algebra probs

Forgot the boundary, thanks.