- #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) = y2+3x4-4x2y , 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) = y2+3x4-4x2y and its critical point (0,0):
fx = -8xy+12x3
fy = -4x2+2y
Hence, fx = fy = 0 implies:
Critical point or extremum is (0,0)
---------------------------------------------------------------------------------
METHOD 1:
Now, we check whether (0,0) is a maxima, a minima or a saddle point:
fxx = -8y + 36x2, hence fxx(0,0) = 0
fyy = 2, hence fyy(0,0) = 2
fxy = fyx = -8x, hence fxy(0,0) = 0
Now, we apply the second derivative test for maxima-minima:
fxxfyy – (fxy)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) = m2x2 + 3x4 – 4mx3
g’(x) = 2m2x + 12x3 – 12mx2
g”(x) = 2m2 +36x2 – 24mx
At x=0 (and hence at y=0):
g(0) = 0
g’(0) = 0
g”(0) = 2m2
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) = 3x4
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.
---------------------------------------------------------------------------------
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 = 2x2 in the function f(x,y):
So, f(x,2x2) = -x4.
On the x-y plane, y = 2x2 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 = 2x2, (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 1012 looks like this:
The graph was drawn using Mathematica 7.0 and the following command was used:
Plot3D[1012 (y2+3x4-4y x2),{x,0.002,-0.002},{y,2*10-6,-2*10-6},PlotRange->{-2,2}]
The value of 1012f(±0.001,0.000002) = -1, calculated using mathematica.
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) = y2+3x4-4x2y , 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) = y2+3x4-4x2y and its critical point (0,0):
fx = -8xy+12x3
fy = -4x2+2y
Hence, fx = fy = 0 implies:
Critical point or extremum is (0,0)
---------------------------------------------------------------------------------
METHOD 1:
Now, we check whether (0,0) is a maxima, a minima or a saddle point:
fxx = -8y + 36x2, hence fxx(0,0) = 0
fyy = 2, hence fyy(0,0) = 2
fxy = fyx = -8x, hence fxy(0,0) = 0
Now, we apply the second derivative test for maxima-minima:
fxxfyy – (fxy)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) = m2x2 + 3x4 – 4mx3
g’(x) = 2m2x + 12x3 – 12mx2
g”(x) = 2m2 +36x2 – 24mx
At x=0 (and hence at y=0):
g(0) = 0
g’(0) = 0
g”(0) = 2m2
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) = 3x4
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.
---------------------------------------------------------------------------------
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 = 2x2 in the function f(x,y):
So, f(x,2x2) = -x4.
On the x-y plane, y = 2x2 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 = 2x2, (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 1012 looks like this:
The graph was drawn using Mathematica 7.0 and the following command was used:
Plot3D[1012 (y2+3x4-4y x2),{x,0.002,-0.002},{y,2*10-6,-2*10-6},PlotRange->{-2,2}]
The value of 1012f(±0.001,0.000002) = -1, calculated using mathematica.