- #1

shanu_bhaiya

- 64

- 0

**The Problem:**

Problem is reference to Problem 23 on Page 210 of Basic Multivariable Calculus by Marsden, Tromba, Weinstein. But my problem is a bit different.

The main problem is:

**For the function f(x,y) = y**

^{2}+3x^{4}-4x^{2}y , is (0,0) a local maxima, minima or saddle point?The secondary problem is that after the failure of method 1, the methods 2 and 3 contradict and method 4 supports method 3.

I approach the problem via different methods given below, some of which prove that (0,0) is a minima and others prove that it is maxima.

**The Attempt at a Solution:**

We investigate the function f(x,y) = y

^{2}+3x

^{4}-4x

^{2}y and its critical point (0,0):

f

_{x}= -8xy+12x

^{3}

f

_{y}= -4x

^{2}+2y

Hence, f

_{x}= f

_{y}= 0 implies:

Critical point or extremum is (0,0)

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**METHOD 1:**

Now, we check whether (0,0) is a maxima, a minima or a saddle point:

f

_{xx}= -8y + 36x

^{2}, hence f

_{xx}(0,0) = 0

f

_{yy}= 2, hence f

_{yy}(0,0) = 2

f

_{xy}= f

_{yx}= -8x, hence f

_{xy}(0,0) = 0

Now, we apply the second derivative test for maxima-minima:

f

_{xx}f

_{yy}– (f

_{xy})

^{2}

= 0 x 2 – 0 = 0.

Hence, second derivative test fails in this case.

---------------------------------------------------------------------------------

Hence, we think of an alternative method as following:

**METHOD 2:**

We put y = mx in the equation and then judge the function according to the one-variable calculus:

f(x,mx) = g(x) = m

^{2}x

^{2}+ 3x

^{4}– 4mx

^{3}

g’(x) = 2m

^{2}x + 12x

^{3}– 12mx

^{2}

g”(x) = 2m

^{2}+36x

^{2}– 24mx

At x=0 (and hence at y=0):

g(0) = 0

g’(0) = 0

g”(0) = 2m

^{2}

Hence g”(0) is positive for all m≠0. Hence, from whichever direction we approach except from x-axis, (0,0) is a minima.

For m=0, the test fails. So, we handle it separately:

When m=0, i.e. f(x,0) = 3x

^{4}

We know that (0,0) is an absolute minima for this function.

Hence, (0,0) is a local minima for f(x,y) from this method.

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Now, we prove that (0,0) is surrounded by some negative points which contradicts the definition of local minima:

**METHOD 3:**

It’s simple, put y = 2x

^{2}in the function f(x,y):

So, f(x,2x

^{2}) = -x

^{4}.

On the x-y plane, y = 2x

^{2}is a parabola. Hence, all the values of the function f(x,y) on this parabola are negative. That means if we approach (0,0) while running on this parabola, we will always encounter negative values except at (0,0) where it is 0, hence on y = 2x

^{2}, (0,0) is a maxima (absolute, infact).

Another important fact is that function at (0,0) is surrounded by some negative values, which contradicts the definition of local minima for a function in this case.

---------------------------------------------------------------------------------

**METHOD 4:**

I thought that going numerically and graphically will somewhat solve my problem. Here are some of my efforts:

The graph of f(x,y) (x from -0.002 to 0.002 and y from -0.000002 to 0.000002) amplified by 10

^{12}looks like this:

The graph was drawn using Mathematica 7.0 and the following command was used:

Plot3D[10

^{12}(y

^{2}+3x

^{4}-4y x

^{2}),{x,0.002,-0.002},{y,2*10

^{-6},-2*10

^{-6}},PlotRange->{-2,2}]

The value of 10

^{12}f(±0.001,0.000002) = -1, calculated using mathematica.