# Thread: Maximum/Minimum

1. ## Maximum/Minimum

It's all in the pdf.

10x in advance.

2. do you know the second derivative test? if so, you can do it..

3. Originally Posted by kalagota
do you know the second derivative test? if so, you can do it..
What's that?

4. Originally Posted by asi123
What's that?
If the second derivative is positive at a critical value then the critical value is a relative minimum. If the second derivative is negative then the critical value is a relative maximum.

-Dan

EDIT: Sorry, I hadn't looked at the actual problem. The test I'm referring to is for one dimensional Calculus. I'll let someone who knows the subject better take this one.

5. Originally Posted by topsquark
If the second derivative is positive at a critical value then the critical value is a relative minimum. If the second derivative is negative then the critical value is a relative maximum.

-Dan

Ok, here is the second derivative, how can I know if it's positive or negative for a given point?

What do I do with dx and dy?

10x.

6. Ok, I couldn't open your PDF and your image won't appear either.

But here is what I think you are asking

Let $(a,b)$ denote a critical point of $f(x,y)$

In other words $f_x(a,b)=f_y(a,b)=0$

or $f_x(a,b)\quad\text{or}\quad{f_y(a,b)}$ is undefined

Now define

$D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-\left\{f_{xy}(a,b)\right\}^2$

Now if $D(a,b)>0\quad\text{and}\quad{f_{xx}(a,b)>0}$ then the point $(a,b)$ is a relative min.

If $D(a,b)>0\quad\text{and}\quad{f_{xx}(a,b)<0}$ then $(a,b)$ is a relative max.

If $D(a,b)<0$ then $(a,b)$ is a Saddle Point of $f(x,y)$

And if $D(a,b)=0$ no conclusion can be drawn.

Is this what you are asking about? If not just respond saying so.

Mathstud.

7. Originally Posted by Mathstud28
Ok, I couldn't open your PDF and your image won't appear either.

But here is what I think you are asking

Let $(a,b)$ denote a critical point of $f(x,y)$

In other words $f_x(a,b)=f_y(a,b)=0$

or $f_x(a,b)\quad\text{or}\quad{f_y(a,b)}$ is undefined

Now define

$D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-\left\{f_{xy}(a,b)\right\}^2$

Now if $D(a,b)>0\quad\text{and}\quad{f_{xx}(a,b)>0}$ then the point $(a,b)$ is a relative min.

If $D(a,b)>0\quad\text{and}\quad{f_{xx}(a,b)<0}$ then $(a,b)$ is a relative max.

If $D(a,b)<0$ then $(a,b)$ is a Saddle Point of $f(x,y)$

And if $D(a,b)=0$ no conclusion can be drawn.

Is this what you are asking about? If not just respond saying so.

Mathstud.

I rescan everything in BMP so I hope you can see it now.

What I'm talking about is a function with a condition, after I find all the points, how do I determinate which one is maximum and which one is minimum, I know I need to use d^2f somehow but I dont know how.

10x in advance.

8. Originally Posted by asi123
I rescan everything in BMP so I hope you can see it now.

What I'm talking about is a function with a condition, after I find all the points, how do I determinate which one is maximum and which one is minimum, I know I need to use d^2f somehow but I dont know how.

10x in advance.
Im really sorry, but once again I cannot see it. If you are not talking about finding relative extrema then I assume this is a question of either Absolute Extrema in a bounded region or a question of Lagrange Multipliers.

Now this may sound novel, but have you tried just evaluating all the points, and from there since you know that they are all relative extrema you can seee which are the absolute max and mins?

9. Originally Posted by Mathstud28
Im really sorry, but once again I cannot see it. If you are not talking about finding relative extrema then I assume this is a question of either Absolute Extrema in a bounded region or a question of Lagrange Multipliers.

Now this may sound novel, but have you tried just evaluating all the points, and from there since you know that they are all relative extrema you can seee which are the absolute max and mins?

You right, I'm talking about Lagrange Multipliers, lets say I have a function f(x,y) and also a condition g(x,y), and I want to build a function l(x,y) = f(x,y) + gama * g(x,y). From here I find all the points, my question is how can I tell which one is minimum and which one is maximum? can I just put them in f(x,y) and evaluate it from there? shouldn't I use d^2l or something like that?

10x again.

10. Ok, so I finally see your page it says

Maximize $f(x,y)=x^2+12xy+2y^2$

Under the condition that

$4x^2+y^2=25$

So defining

$g(x,y)=4x^2+y^2-25$

We need to solve the following system of equations

$\left\{\begin{array}{cc}f(x,y)_x=\lambda{g(x,y)_x} \\f(x,y)_y=\lambda{g(x,y)_y}\\g(x,y)=0\end{array}\ right\}$

So computing

$f_x=2x+12y$

$f_y=12x+4y$

$g_x=8x$

$g_y=2y$

We see that we must solve

$\left\{\begin{array}{cc}2x+12y=8\lambda{x}\\12x+4y =2\lambda{y}\\4x^2+y^2-25=0\end{array}\right\}$

So we see that the four solutions are

$\left\{\begin{array}{c}x=-2\\y=3\\z=-2\end{array}\right\}$

and

$\left\{\begin{array}{c}x=\frac{-3}{2}\\y=-4\\z=\frac{17}{4}\end{array}\right\}$

and

$\left\{\begin{array}{c}x=\frac{3}{2}\\y=4\\z=\frac {17}{4}\end{array}\right\}$

and

$\left\{\begin{array}{c}x=2\\y=-3\\z=-2\end{array}\right\}$

So testing we get

$f(-2,3)=-50$

$f\left(\frac{-3}{2},-4\right)=\frac{425}{4}$

$f\left(\frac{3}{2},4\right)=\frac{-151}{4}$

$f(2,-3)=-50$

You can draw your conclusions from there.