Need help in finding extreme values

[tiny-tim]

Not so fast. -(x-y)² ≤ 0. The possibility of 0 means we have not yet established whether O is a maximum or a saddle. To resolve this, you have to look further down the Taylor expansion.

not exactly … our quadratic formula gives all the information we need except in one direction (which we can check "by hand"):

0 is a possibility only along x = y

so, near (0,0), our quadratic formula, -4(x-y)², tells us that there is a maximum along any direction other than x = y

however, if gives us no information about x = y, so that might be a maximum also, or it might not … we need a cubic or quartic formula to check, or we can check "by hand"

(we could have got all this by simply putting all the lines x = constant*y …
obviously the x⁴ + y⁴ is insignificant, and the -2(x-y)² dominates (unless the constant = 0), so clearly there is a maximum in every direction except that there is a minimum along x = y

the surface starts to go down everywhere, except that it goes slowly up (~ r⁴) along x = y )​

… we have the awkward situation that at O it is only defined in the second and fourth quadrants …

no, it's defined everywhere except between a couple of curves that approach O tangentially to x = y

 

[aruwin]

tiny-tim, can you help me with points A and B? I have expanded the function and I am stucked =(

 

[tiny-tim]

… I have expanded the function and I am stucked =(

(you mean "stuck" )

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

 

[aruwin]

(you mean "stuck" )

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

Isn't the f_xx formula is derived from the taylor expansion??I'll copy it again here =)

The taylor expansion at A(√2,-√2) is
f(x,y)≈ -8 + 1/2![20(x-√2)2+8(x-√2)(y+√2)+20(y+√2)2]+...

The taylor polynomial until the 2nd term is
f(x,y) = -8+10(x-√2)2+4(x-√2)(y+√2)+10(y+√2)2

I think I have figured something out.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]

not exactly … our quadratic formula gives all the information we need except in one direction (which we can check "by hand"):

I'd understood from earlier posts that the purpose of the exercise was to practise use of Taylor expansions. You can resolve the ambiguity by going further down the expansion, but in the present case, as you and I have both indicated, it's easier 'by hand'.

no, it's defined everywhere except between a couple of curves that approach O tangentially to x = y

ah yes... it's defined in a figure-of-8, which does spill into the first and third quadrants.

Yes, without the ≤0, O is a saddle. The ≤0 is there to state the region of f(x,y).

OK, but on that argument the whole of the curve along which f = 0 constitutes a local maximum.

I'm stuck here, from the equation, it seems that at (√2,-√2), f(x,y) is always going to be -8 but if I put other values it is always positive or can I just ignore the "-8"?

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.

 

[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)