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

## Homework Statement

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

f (x, y) = x

x2 + 2y2 = 3

## Homework Equations

det = fxx * fyy - (fxy)2
If det > 0,
fxx < 0 $$\Rightarrow$$ MAXIMUM
fxx > 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 = $$\pm$$$$\sqrt{3 - 2y2}$$

Therefore,

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

At critical point,

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

= (0, 0)

Therefore, equating gives

0 = 0

&

- $$\frac{2y}{\sqrt{3 - 2y2}}$$ = 0

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

fxx = 0
fxy = 0

det = fxx * fyy - (fxy)2
det = 0 * fyy - (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...

Related Calculus and Beyond Homework Help News on Phys.org
Mark44
Mentor

## Homework Statement

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

f (x, y) = x

x2 + 2y2 = 3

## Homework Equations

det = fxx * fyy - (fxy)2
If det > 0,
fxx < 0 $$\Rightarrow$$ MAXIMUM
fxx > 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 = $$\pm$$$$\sqrt{3 - 2y2}$$
Tip: Don't mix [ sup] or [ sub] tags inside [ tex] tags. Instead use ^{} for exponents and _{} for subscripts.
Therefore,

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

At critical point,

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

= (0, 0)

Therefore, equating gives

0 = 0
I don't see how the equation above is helpful at all.
&

- $$\frac{2y}{\sqrt{3 - 2y2}}$$ = 0

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

fxx = 0
fxy = 0

det = fxx * fyy - (fxy)2
det = 0 * fyy - (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...

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.

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) e1-x2-y2

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

fx = 2xe1-x2-y2 ( 1 - x2 - 3y2 )
fy = 2ye1-x2-y2 (3 - x2 - 3y2 )

At critical points, (fx, fy) = (0, 0)

Equating gives,

2xe1-x2-y2 ( 1 - x2 - 3y2 ) = 0 --- (1)
2ye1-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.

Min at (0, 0),
max at (0, 1), (0, -1),
saddle at (-1, 0), (1, 0)

Thanks!

vela
Staff Emeritus
Homework Helper
Homework Statement
Find local minima, maxima and saddle points of the function:

f(x, y) = (x2 + 3y2) e1-x2-y2

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

fx = 2xe1-x2-y2 ( 1 - x2 - 3y2 )
fy = 2ye1-x2-y2 (3 - x2 - 3y2 )

At critical points, (fx, fy) = (0, 0)

Equating gives,

2xe1-x2-y2 ( 1 - x2 - 3y2 ) = 0 --- (1)
2ye1-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.

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.

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
Staff Emeritus
Homework Helper
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.