Need help in finding extreme values

[aruwin]

For the purpose of asking whether it's a max etc., the -8 is irrelevant. You don't care what the value of f is there, you only care how the value changes as you move from that point in different directions. If it increases no matter which way you move it's a minimum.

Yes,I was a lil confused before.Now I understand it.
From the quadratic equation that we got from the taylor expansion,-8 is the constant and we have a square and thus, f(x,y)is always going to be bigger or equals to -8 and so we have a minimum value of -8.Am I right?

 

[haruspex]

From the quadratic equation that we got from the Taylor expansion,-8 is the constant and we have a square and thus, f(x,y) is always going to be bigger or equals to -8 and so we have a minimum value of -8. Am I right?

Yes, except that as I mentioned you do have to be careful about the 'or equals' bit.
If there is some path from the point in question along which the first derivative of f is zero then you need to look at the second derivative, and if that is zero then the third derivative, and so on. But at the point (√2, -√2), you don't have that complication. Writing u and v for the x and y distances from that point, the first few terms give you -8 + 8u² + 2(u+v)² + 8v². Any movement away from u = v = 0 will initially increase the value.

 

[aruwin]

Yes, except that as I mentioned you do have to be careful about the 'or equals' bit.
If there is some path from the point in question along which the first derivative of f is zero then you need to look at the second derivative, and if that is zero then the third derivative, and so on. But at the point (√2, -√2), you don't have that complication. Writing u and v for the x and y distances from that point, the first few terms give you -8 + 8u² + 2(u+v)² + 8v². Any movement away from u = v = 0 will initially increase the value.

Since if u=v=0,f(x,y) is -8 and that's why I say f(x,y)is equals to or more than f(sqrt(2),-sqrt(2)). Isn't it so?

 

[haruspex]

Since if u=v=0,f(x,y) is -8 and that's why I say f(x,y)is equals to or more than f(sqrt(2),-sqrt(2)). Isn't it so?

Yes, but you're missing a key detail. It is not enough that f(x,y) is equal to or exceeds -8 in a neighbourhood of the point. You need that it actually exceeds -8 as soon as you move away from that point.

 

[tiny-tim]

Since if u=v=0,f(x,y) is -8 and that's why I say f(x,y)is equals to or more than f(sqrt(2),-sqrt(2)). Isn't it so?

aruwin, you're missing the point

(you're also using bad terminology: that expression is not f, it is an approximation to f, so you should give it a different letter, to show the examiner you understand what you're talking about: i'll use "g" )

at A, g(A + (x,y)) = -8 + 8x² + 2(x+y)² + 8y²

so for all (x,y) ≠ (0,0), g(A + (x,y)) > f(A), so g tells us there is certainly a minimum of f at A

at O, the difficulty was that g(O + (x,y)) = f(O) even for some (x,y) ≠ 0, so g tells us there may be a maximum at O, but we need to check further

 

[aruwin]

aruwin, you're missing the point

(you're also using bad terminology: that expression is not f, it is an approximation to f, so you should give it a different letter, to show the examiner you understand what you're talking about: i'll use "g" )

at A, g(A + (x,y)) = -8 + 8x² + 2(x+y)² + 8y²

so for all (x,y) ≠ (0,0), g(A + (x,y)) > f(A), so g tells us there is certainly a minimum of f at A

at O, the difficulty was that g(O + (x,y)) = f(O) even for some (x,y) ≠ 0, so g tells us there may be a maximum at O, but we need to check further

Oh ok,lets use g for the approximation,then But first of all,the approximation I got at point A is
f(x,y) ≈ -8 + [10(x - √2)^2 + 4(x - √2)(y + √2) + 10(y + √2)^2]
so,
g(x,y) = -8 + [10(x - √2)^2 + 4(x - √2)(y + √2) + 10(y + √2)^2].

This shows that g always increases above -8,right? But if I substitute (√2,-√2) into (x,y), g becomes -8. That's why I thought that the approximation g(x,y)≥f(A) and isn't this the condition to get a local minimum point? And is it correct that I say -8 is the minimum value at A?

 

[tiny-tim]

g(x,y) = -8 + [10(x - √2)^2 + 4(x - √2)(y + √2) + 10(y + √2)^2]

(i haven't checked that, but …)

let's rewrite that as:

g(A + (x,y)) = f(A) + 10x² + 4xy + 10y²

then you need to prove that the 10,4,10 part is always strictly positive …

how would you do that? ​

 

[aruwin]

(i haven't checked that, but …)

let's rewrite that as:

g(A + (x,y)) = f(A) + 10x² + 4xy + 10y²

then you need to prove that the 10,4,10 part is always strictly positive …

how would you do that? ​

By completing the square??

 

[tiny-tim]

yup!

 

[aruwin]

yup!

Ok,this is what I got:

g(x,y) = -8 + 10 {[(x - √2)^2 + (1/5)(y + √2)]^2 + (24/25)(y + √2)^2].

Now it's proven that g(x,y) is ≥f(A) ?? And the the minimum value is -8?

 

[tiny-tim]

g(x,y) = -8 + 10 {[(x - √2)^2 + (1/5)(y + √2)]^2 + (24/25)(y + √2)^2]

but there's no xy in that!

10x² + 4xy + 10y²

= 10(x-y)² + 24xy = 10(x+y)² - 16xy

so it lies strictly between 10(x-y)² and 10(x+y)², and so must be positive (unless x= = y = 0)

(alternatively, use the determinant)

 

[aruwin]

but there's no xy in that!

10x² + 4xy + 10y²

= 10(x-y)² + 24xy = 10(x+y)² - 16xy

so it lies strictly between 10(x-y)² and 10(x+y)², and so must be positive (unless x= = y = 0)

(alternatively, use the determinant)

Actually,I was completing the square of the taylor polynomial which I stated earlier.
That is,
f(x,y) ≈ -8 + [10(x - √2)^2 + 4(x - √2)(y + √2) + 10(y + √2)^2]

to this:
-8 + 10 {[(x - √2)^2 + (1/5)(y + √2)]^2 + (24/25)(y + √2)^2]

So, now it is positive. Can I say that g(x,y)≥f(A) and thus,it has a minimum value of -8?

 

[haruspex]

-8 + 10 {[(x - √2)^2 + (1/5)(y + √2)]^2 + (24/25)(y + √2)^2]

So, now it is positive. Can I say that g(x,y)≥f(A) and thus,it has a minimum value of -8?

That's all fine except that g(x,y)≥f(A) is not strong enough to make it a minimum. You need g(x,y)>f(A) for all (x,y) ≠ A (in some neighbourhood). This is because g is not f - it's only a local approximation to f (which I'll refer to as g_A hereon). If there were some path through A along which g_A(x,y) = f(A) then later terms in the Taylor expansion might result in f decreasing along that path. A good example is what happens to this same function at O (if you take away the f ≤ 0 constraint). The Taylor series down to the second derivatives gives g_O ≤ f(O), but it's a saddle, not a max.
The expression for g_A as a sum of squares does give you what you need. It is stronger than merely g_A(x,y)≥f(A).

 

[HallsofIvy]

(you mean "stuck" )

but doesn't your professor want you to use the f_xx etc formula?

If the problem is to find the global max and min on a set, then the second derivative formula, which determines whether a critical point gives a local max, min, or saddle point, is irrelevant. Determine all possible points at whicy max or min can occur, where the gradient is 0 in the interior of the set, or on the boundary, calculate the function values at those points, and see which are max and min.

 

[aruwin]

That's all fine except that g(x,y)≥f(A) is not strong enough to make it a minimum. You need g(x,y)>f(A) for all (x,y) ≠ A (in some neighbourhood). This is because g is not f - it's only a local approximation to f (which I'll refer to as g_A hereon). If there were some path through A along which g_A(x,y) = f(A) then later terms in the Taylor expansion might result in f decreasing along that path. A good example is what happens to this same function at O (if you take away the f ≤ 0 constraint). The Taylor series down to the second derivatives gives g_O ≤ f(O), but it's a saddle, not a max.
The expression for g_A as a sum of squares does give you what you need. It is stronger than merely g_A(x,y)≥f(A).

Thanks for all your explanations and patience.Yes,I have solved this question with your help and harupex too.By the wsy,I have a new question.Since it is related to finding local extrema too,I am just going to ask it here.
Given the equation x^2+y^2+3z^2-2xy-2yz=2.Find the local extrema for the function z=f(x,y).What do I do with the equation?

 

[haruspex]

Given the equation x^2+y^2+3z^2-2xy-2yz=2.Find the local extrema for the function z=f(x,y).What do I do with the equation?

What do you get when you take the partial derivatives?

 

[aruwin]

What do you get when you take the partial derivatives?

Is this ok?
Partial derivative in respect to x is 2x-2y and in respect to y is 4y-2x-2z.
What do I do with z??

 

[HallsofIvy]

Thanks for all your explanations and patience.Yes,I have solved this question with your help and harupex too.By the wsy,I have a new question.Since it is related to finding local extrema too,I am just going to ask it here.
Given the equation x^2+y^2+3z^2-2xy-2yz=2.Find the local extrema for the function z=f(x,y).What do I do with the equation?

You want to differentiate z with respect to x and y and can do that using "implicit differentiation" just like you learned in Calculus I.

 

[haruspex]

Is this ok?
Partial derivative in respect to x is 2x-2y and in respect to y is 4y-2x-2z.
What do I do with z??

Partial derivative of z wrt x is ∂z/∂x, etc. Just treat that as a variable. You can differentiate it partially wrt x again, ∂²z/∂x², or wrt y: ∂²z/∂x∂y. (Btw, ∂²z/∂x∂y = ∂²z/∂y∂x)