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**Karimazer1****Guest**

Hey! It's been a while! Hope you are all good!

I have simple thing I cannot get my head around.

So far; the function that we want to maximise is x1^2/2 +x1x2.

Differentiated, gives x1+x2.

SOC is x2.

1. General question - if the SOC = 0...what does that mean? I.E When SOC<0 =max, SOC>0 =min, what do you say when SOC=0?

2. x1 AND x2 can be -1, 0 or 1. We want to find the maximum. Is the following correct?

IF x2 = 1, x1 will be maximised when x1=1, but SOC is 1 (>0) therefore, this is a minimum...so it is not a maximum?

IF x2 = -1, x1 will be maximised when x1=1 too, now SOC is -1 <0 therefore maximum!

Please help me answer the first general question and let me know if my working/logic behind the second question is correct ! Thank you !!!!

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,258

Hi;

Welcome back!

SOC?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Karimazer1****Guest**

Second order condition, sorry.

I don't know what is wrong with me, it seems my basic basic maths skills have gone. I've just...forgotten?

I mean (in addition to the above!) when you differentiate x1+x2 with respect to x1, you do get x2, right? And not 1+x2?

Same applies the other way, with x2-x1, differentiate with respect to 2, we get -x1 right? not 1-x1?

Thank you !!

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,428

hi Karimazer1

I had lots of trouble under standing your post.

Here's what I'm getting for your first question.

(i) You've differentiated to find stationary points.

(ii) you've differentiated a second time so you can say max or min.

But what to do if the second differential is zero?

Answer you cannot easily tell I'm afraid.

Consider these graphs all with d2y/dx^2 = 0

y = x^3 This has a point of inflexion at x = 0

y = x^4 This has a minimum at x = 0

y - -x^4 This has a maximum at x = 0

I seem to vaguely remember somebodies rule that you keep differentating until you don't get a zero but I'm uncertain. (have to do some research)

Now to the second question.

Are you treating x1 and x2 as independent variables and you are using partial differentiation ?

Because now we have other possibilities. As well as max and min you can have saddle points (max in one plane but min in the other) or even max for one variable but no stationary point for the other variable.

Whatever, I don't follow what your function is for Q2.

Suggest you start by defining this more clearly.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

Hi

http://en.wikipedia.org/wiki/Higher-order_derivative_test

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**Karimazer1****Guest**

Ok can we please start simply?

If:

x1/2 + x1x2

What is it differentiated?

Then what is that differentiated again?

It is; 1? Right?

Thanks

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,428

hi Karimazer1

If

The deltas are used in place of 'd' to show that one variable is being treated as a constant while differentiation is applied to the other.

On a 3D graph, the result is as if you had cut through the surface with a plane parallel to one of the x axes.

To come back to your question, which variable are you differentiating with respect to ?

Bob'

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Karimazer1****Guest**

Hi bob thank you. I mean

If y = (x1)/2 + x1*x2

So the SOC is still 1? Right? Or am I wrong?

**Karimazer1****Guest**

Diferentiating with respect to x1,

there are two variables, x1 and x2

y = x1/2 +x1x2

first time = x1 + x2

second time is 1?

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,428

hi

Looking back I see you had a different function

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Karimazer1****Guest**

Thank you Bob so much!

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