two calculus problems: classify critical points of f(x,y)...

[laovienax]

I am taking a math crash-course for an economics degree and am catching up with a lot of stuff I should have learned long ago. I'm having some difficulties with my current problem set. If someone could quickly look over it, I'd really appreciate it. I am not looking for exact solutions; some hints on where I went wrong or what I should do or look up would be very helpful.

The instructions for the questions are as follows:

For each of the following functions, find the critical points and classify these as local max, local min, or 'can't tell':

1) f(x, y) = x^4 + x^2 - 6xy + 3y^2

Is it possible that this function has two minima and one saddlepoint?

I get three critical points, two of which have both Hessian principal minors greater zero (=minimum) and one which has the first principal minor positive but the second negative, which should be a saddlepoint. But how could this be possible? If I have two minima, don't I need a maximum somewhere inbetween?

My calculations are:

. . .first order conditions:

. . .d f(x,y)/dx = 4x^3 + 2x - 6y = 0

. . .d f(x,y)/dy = -6x + 6y = 0

I substracted one from the other to get rid of the 6y, so I end up with x(4x^2 - 4) = 0 from which i got the three critical points x1= (0, 0); x2= (1, 1); x3=(-1, -1). Taking partial derivatives again for the second order conditions, I get:

. . .d^2 f(x,y)/dx^2 = 12x^2 + 2
. . .d^2 f(x,y)/dxdy = -6
. . .d^2 f(x,y)/dydx = -6
. . .d^2 f(x,y)/dy^2 = 6

This yields the two hessian principal minors:

. . .H1=12x^2 + 2
. . .H2 = 72x^2 - 24

This in turn yields the above mentioned results if you substitute the critical points in. But is this correct? And how can I check whether these minima are GLOBAL minima?

My second question concerns another of these functions, but i got stuck very early on; the function is:

2) f(xy) = xy^2 + x^3y - xy

If I take first order derivatives, I get:

. . .f'/dx= y^2 + 3x^2y - y = 0
. . .f'/dy = 2xy + x^3 - x = 0

But I don't see how to get to the critical points. How do I get rid of either x or y in one of the equations?

Thank you so much for your help!
lv

 

[galactus]

For #2:

If you take the first derivatives of \(\displaystyle \L\\xy^{2}+x^{2}y-xy\)

You get: \(\displaystyle \L\\f_{x}(x,y)=3x^{2}y+y^{2}-y\)

\(\displaystyle \L\\f_{y}(x,y)=2xy+x^{3}-x\)

The critical points satisfy:

\(\displaystyle \L\\3x^{2}y+y^{2}-y=0\)........[1]

\(\displaystyle \L\\2xy+x^{3}-x=0\).........[2]

Solve [1] for \(\displaystyle y=1-3x^{2}\;\ or\;\ 0\)

Sub into [2] and solve for x(you could do other solvings and subbings):

\(\displaystyle \L\\2x(1-3x^{2})+x^{3}-x=0\)

\(\displaystyle x=\pm\frac{1}{\sqrt{5}},\;\ 0\)

This yields \(\displaystyle \L\\\left(\frac{1}{\sqrt{5}},\frac{2}{5}\right),\;\ \left(\frac{-1}{\sqrt{5}},\frac{2}{5}\right),\;\ \left(0, 1\right)\)


Now, we can test the critical points using the second partial test:

\(\displaystyle \L\\D=f_{xx}(x,y)f_{yy}(x,y)-f^{2}_{xy}(x,y)\)......[3]

If \(\displaystyle D>0\;\ and\;\ f_{xx}(x,y)>0\), then relative minimum at (x,y)

If \(\displaystyle D>0\;\ and \;\ f_{xx}<0\), then relative maximium at f(x,y)

If \(\displaystyle D<0\), then f has a saddle point at (x,y)

If D=0, then no conclusion can be made.

Let's see what you have:

\(\displaystyle f_{xx}(x,y)=6xy\)

\(\displaystyle f_{yy}(x,y)=2x\)

\(\displaystyle f_{xy}(x,y)=2y+3x^{2}-1\)

Now, test your points using the formula in [3]

 

[laovienax]

thank you