Maxima/Minima of extrema function f(x,y)=x

[RyanV]

Homework Statement

Find the extrema of the function subject to the given constraint.

f (x, y) = x

x² + 2y² = 3

Homework Equations

det = f_xx * f_yy - (f_xy)²
If det > 0,
f_xx < 0 $$\Rightarrow$$ MAXIMUM
f_xx > 0 $$\Rightarrow$$ MINIMUM

If det < 0,
$$\Rightarrow$$ SADDLE POINT

If det = 0m
Inconclusive

The Attempt at a Solution

I appear to be struggling with this maxima/minima stuff. Previous questions were just correct. I had to look at the book's answers to figure a way there. I seem to be missing something...but here is my attempt:

Since x² + 2y² = 3,

x = $$\pm$$[tex]\sqrt{3 - 2y²}[/tex]

Therefore,

f(x, y) = $$\pm$$[tex]\sqrt{3 - 2y²}[/tex]

At critical point,

$$\nabla$$ f = (0, - [tex]\frac{2y}{\sqrt{3 - 2y²}}[/tex] )

= (0, 0)


Therefore, equating gives

0 = 0

&

- [tex]\frac{2y}{\sqrt{3 - 2y²}}[/tex] = 0

So extrema points at ($$\sqrt{3}$$, 0) & (-$$\sqrt{3}$$, 0) since y = 0.


To find max/min/saddle, find det.

f_xx = 0
f_xy = 0

det = f_xx * f_yy - (f_xy)²
det = 0 * f_yy - (0)²
det = 0

Which is inconclusive. The book has ($$\sqrt{3}$$, 0) as a maximum and (-$$\sqrt{3}$$, 0) as a minimum.

I don't know what I'm missing out here...

Thanks in advance.

 

[Mark44]

Homework Statement

Find the extrema of the function subject to the given constraint.

f (x, y) = x

x² + 2y² = 3

Homework Equations

det = f_xx * f_yy - (f_xy)²
If det > 0,
f_xx < 0 $$\Rightarrow$$ MAXIMUM
f_xx > 0 $$\Rightarrow$$ MINIMUM

If det < 0,
$$\Rightarrow$$ SADDLE POINT

If det = 0m
Inconclusive

The Attempt at a Solution

I appear to be struggling with this maxima/minima stuff. Previous questions were just correct. I had to look at the book's answers to figure a way there. I seem to be missing something...but here is my attempt:

Since x² + 2y² = 3,

x = $$\pm$$[tex]\sqrt{3 - 2y²}[/tex]

Tip: Don't mix [ sup] or [ sub] tags inside [ tex] tags. Instead use ^{} for exponents and _{} for subscripts.

Therefore,

f(x, y) = $$\pm$$[tex]\sqrt{3 - 2y²}[/tex]

At critical point,

$$\nabla$$ f = (0, - [tex]\frac{2y}{\sqrt{3 - 2y²}}[/tex] )

= (0, 0)


Therefore, equating gives

0 = 0

I don't see how the equation above is helpful at all.

&

- [tex]\frac{2y}{\sqrt{3 - 2y²}}[/tex] = 0

So extrema points at ($$\sqrt{3}$$, 0) & (-$$\sqrt{3}$$, 0) since y = 0.


To find max/min/saddle, find det.

f_xx = 0
f_xy = 0

det = f_xx * f_yy - (f_xy)²
det = 0 * f_yy - (0)²
det = 0

Which is inconclusive. The book has ($$\sqrt{3}$$, 0) as a maximum and (-$$\sqrt{3}$$, 0) as a minimum.

I don't know what I'm missing out here...

Thanks in advance.

What I think you are missing is some understanding of what f(x, y) = x looks like, which is a plane. Since a plane is a flat surface with no curvature, there are no dips or bumps. Also, since the domain of this plane is restricted to {(x, y) | x² + 2y² = 3}, any maximum or minimum points are attained at points the farthest away from the origin.

 

[RyanV]

Ohh...I see... I feel so silly ><

Thanks a lot for that.



If I may, I have another question regarding maxima and minima...

Homework Statement
Find local minima, maxima and saddle points of the function:

f(x, y) = (x² + 3y²) e^1-x2-y2


I started off finding the partial derivatives of x and y respectively and got:

f_x = 2xe^1-x2-y2 ( 1 - x² - 3y² )
f_y = 2ye^1-x2-y2 (3 - x² - 3y² )

At critical points, (f_x, f_y) = (0, 0)

Equating gives,

2xe^1-x2-y2 ( 1 - x² - 3y² ) = 0 --- (1)
2ye^1-x2-y2 (3 - x² - 3y² )= 0 --- (2)


After that I'm kinda lost...I slithered my way to get (0, 0) but very unconvincingly. I would rather have some solid reasoning as to why.

Like I mentioned earlier, I think I'm missing some understanding and maybe that's why I'm not getting this.

The answers give
Min at (0, 0),
max at (0, 1), (0, -1),
saddle at (-1, 0), (1, 0)

Thanks!

 

[vela]

Homework Statement
Find local minima, maxima and saddle points of the function:

f(x, y) = (x² + 3y²) e^1-x2-y2I started off finding the partial derivatives of x and y respectively and got:

f_x = 2xe^1-x2-y2 ( 1 - x² - 3y² )
f_y = 2ye^1-x2-y2 (3 - x² - 3y² )

At critical points, (f_x, f_y) = (0, 0)

Equating gives,

2xe^1-x2-y2 ( 1 - x² - 3y² ) = 0 --- (1)
2ye^1-x2-y2 (3 - x² - 3y² )= 0 --- (2) After that I'm kinda lost...I slithered my way to get (0, 0) but very unconvincingly. I would rather have some solid reasoning as to why.

Like I mentioned earlier, I think I'm missing some understanding and maybe that's why I'm not getting this.

The answers give
Min at (0, 0),
max at (0, 1), (0, -1),
saddle at (-1, 0), (1, 0)

When you have a product equal to 0, then one of the factors must be equal to 0. For instance, from this equation

$$2xe^{1-x^2-y^2}(1-x^2-3y^2) = 0$$

you can conclude that x=0 or 1-x²-3y²=0. Similarly, the other equation gives you y=0 or 3-x²-3y²=0. Now you just have to find the solutions for all four possible combinations:

(1) x=0, y=0;
(2) x=0, 3-x²-3y²=0;
(3) y=0, 1-x²-3y²=0; and
(4) 1-x²-3y²=0, 3-x²-3y²=0.

 

[RyanV]

Ooo, I see now. I think I managed to confuse myself by using how for eq (1),

since x = 0,

1-x²-3y²=0
1 - 3y² = 0

which is definitely not looking good in terms of the answers. But I understand now =)

One last thing, so for the 4th solution: (4) 1-x²-3y²=0, 3-x²-3y²=0.

There isn't a solution for this yeah? Since all the variables x/y cancel out.

 

[vela]

Ooo, I see now. I think I managed to confuse myself by using how for eq (1),

since x = 0,

1-x²-3y²=0
1 - 3y² = 0

Right idea, but wrong equation. That's the one that has to hold if y is 0 and x isn't 0.

One last thing, so for the 4th solution: (4) 1-x²-3y²=0, 3-x²-3y²=0.

There isn't a solution for this yeah? Since all the variables x/y cancel out.

Yup, that combination has no solutions.