# Thread: Extrema (Absolute) - Two variables

1. ## Extrema (Absolute) - Two variables

Given the following function, determine the absolute extrema and find where they occur.

f(x,y) = 2x^4 + y^3 - x^2y on the region that is bounded by y = -1 and y = 3 - x^2

WORK:

The first thing I think to do is find the gradient.

I don't suppose disciminants could be useful, as I think they only apply to LOCAL extrema.....

That is, D(a,b) = f_(xx)(a,b)*f_(yy)(a,b) - [f_(xy)(a,b)]^2, where f_(x), for example, is the partial with respect to x, and then xx is then the partial with respect to x of the one we just found, etc. Then, if D(a,b) > 0 and f_(xx)(a,b) > 0, then f has a local min. at (a,b)... if D(a,b) > 0 and f_(xx)(a,b) < 0, blah blah blah, if D(a,b) < 0, then f has a saddle point at (a,b), and so on. But again, I think this only applies to local.

So then, I think of trying to look at this as I would in Calc 1 finding extrema. That is, find critical points, test those, and test endpoints (which I will assume is the boundaries we are given), etc. We can find crit. points by just taking the partial with respect to x, and the partial with respect to y, and plugging them to 0, I believe, just as you would a derivative.

f_(x) = 8x^3 - 2xy
f_(y) = 3x^2 - x^2

So, f_(x) = 8x^3 - 2xy = 0 ... happens when x = 0 or y = 0
f_(y) = 3x^2 - x^2 .. happens when x = 0

This is where things become a little fuzzy.

2. Originally Posted by Ideasman
Given the following function, determine the absolute extrema and find where they occur.

f(x,y) = 2x^4 + y^3 - x^2y on the region that is bounded by y = -1 and y = 3 - x^2
The absolute minimum you need to do this is the following peice
of information:

The global maximum (minimum) of a constrained optimisation problem
is either one of the claculus type local maxima (minima) in the interior
of the feasible region, or occurs on the boundary.

RonL

3. Originally Posted by Ideasman
Given the following function, determine the absolute extrema and find where they occur.

f(x,y) = 2x^4 + y^3 - x^2y on the region that is bounded by y = -1 and y = 3 - x^2

WORK:

The first thing I think to do is find the gradient.

I don't suppose disciminants could be useful, as I think they only apply to LOCAL extrema.....

That is, D(a,b) = f_(xx)(a,b)*f_(yy)(a,b) - [f_(xy)(a,b)]^2, where f_(x), for example, is the partial with respect to x, and then xx is then the partial with respect to x of the one we just found, etc. Then, if D(a,b) > 0 and f_(xx)(a,b) > 0, then f has a local min. at (a,b)... if D(a,b) > 0 and f_(xx)(a,b) < 0, blah blah blah, if D(a,b) < 0, then f has a saddle point at (a,b), and so on. But again, I think this only applies to local.

So then, I think of trying to look at this as I would in Calc 1 finding extrema. That is, find critical points, test those, and test endpoints (which I will assume is the boundaries we are given), etc. We can find crit. points by just taking the partial with respect to x, and the partial with respect to y, and plugging them to 0, I believe, just as you would a derivative.

f_(x) = 8x^3 - 2xy
f_(y) = 3x^2 - x^2

So, f_(x) = 8x^3 - 2xy = 0 ... happens when x = 0 or y = 0
happens when x=0 or 4x^2=y

f_(y) = 3x^2 - x^2 .. happens when x = 0
f_(y) = 2y^2 - x^2

and f_(y)=0, when y=+/-(1/2)x

So x=0, y=0 satisfies f_(x)=0, f_(y)=0, but you also need to check
for other points in the feasible region where y=4x^2, and y=+/-(1/2)x
meet.

Then you will need also to check for extrema on the boundary of the feasible
region.

RonL

4. Originally Posted by CaptainBlack
happens when x=0 or 4x^2=y

f_(y) = 2y^2 - x^2

and f_(y)=0, when y=+/-(1/2)x

So x=0, y=0 satisfies f_(x)=0, f_(y)=0, but you also need to check
for other points in the feasible region where y=4x^2, and y=+/-(1/2)x
meet.

Then you will need also to check for extrema on the boundary of the feasible
region.

RonL

Thanks for the help, Ron.

I realize I got my partial with respect to y wrong (as did you, but I suspect you oversaw it), and it should be:

f_y = 3y^2 - x^2

Going back to f_x..

f_x = 8x^3 - 2xy..

So now we want to see when these partials are equal to 0.

So, 8x^3 - 2xy = 0 --> (4x^3 - xy) = 0 --> equal to zero when:

x = 0 or y = 4x^2, I totally agree.

Now for f_y..

f_y = 3y^2 - x^2 = 0 --> Equal to zero when: I *THINK* it's when x = +/- sqrt(3)*y

You had: "and f_(y)=0, when y=+/-(1/2)x" which doesn't work, but then again you didn't get the same f_y as me.

Any way, once I find the ones for f_y, I find the values that make both f_x and f_y = 0? This is a little confusing. Also, I'm not quite clear on checking the boundaries... but now I see why this one prob. is worth half the homework grade because it takes so much work! And so confusing

5. Ok, I was wrong, for:

f_x = 8x^3 - 2xy = 0, the only time that is equal to zero is when x = 0 or y = 4x^2 (like you said).

Then, for f_y = 3y^2 - x^2 = 0, the only time this is equal to 0 is when:

x = +/- sqrt(3)*y or y = +/- (sqrt(3)*x)/3

I tried following the work on this site:

Absolute Extrema - Calculus - Online Mathematics

Although, I get lost. For critical points, do I use the points that satisfy BOTH equations? Or are they all the values I got when I got the values for when the partial with respect to x and y are equal to 0? Thus far, I would have:

x = 0, y = 4x^2, x = sqrt(3)*y, x = -sqrt(3)*y, y = (sqrt(3)*x)/3, y = -(sqrt(3)*x)/3

Recall, I am finding the abs extrema on the region y = -1 and y = 3 - x^2.

So now do I take those values (x = 0, y = 4x^2, x = sqrt(3)*y, x = -sqrt(3)*y, y = (sqrt(3)*x)/3, y = -(sqrt(3)*x)/3) and plug them into the orig. equation f(x,y) = 2*x^4 + y^3 - x^2*y, and if I do I won't always be able to see values since it will most likely have variables.... it's tricky since the 'points' I have aren't values, instead have variables.. ahhh

So in general, it's:

1.) Find all the critical points of the function that lie in the region D and determine the function value at each of these points.
2.) Find all extrema of the function on the boundary.
3.) The largest and smallest values found in the first two steps are the absolute minimum and the absolute maximum of the function.

I'm assuming the 'boundary' is y = -1, y = 3 - 3x^2? And how do I find the largest/smallest values when I have variables, once again.

6. Attachment shows the feasible region (the region that satisfies the constraints that y>=-1, and y<=3-x^2).

Now the extrema of f(x,y) = 2x^4 + y^3 - x^2y are either calculus type extrema in the
interior of this region where f_x(x,y) = f_y(x,y) =0, or are on the boundary of the region.

So lets look first for these calculus type extrema.

f_x(x,) = 8x^3 - 2xy = 0

f_y(x,y) = 3y^2 - x^2 = 0

Now if x != 0, the first of these tells us that y=4x^2, substituting this into the second gives:

3 y^2 - y/4 = 0,

so y=0, or y=1/12, but if y=0 we have x=0 and f(x,y)=0, and for the second we have x=sqrt(1/48)
and f(x,y)= 7/(4.12^3).

So we have two critical points in the interior of the feasible region (0,0) and (sqrt(1/48), 1/12).

Now we need to classify these points.

First look at (0,0) look at how f(x,y) varies as we approch (0,0) along the axis x=0, then
f(0,y)=y^3, so y=0 is a point of inflection along this route, so (0,0) is neither a local
maximum or minimum.

Now look at (sqrt(1/48), 1/12), this must be a local minimum, as f(x,y) is smooth, and goes to
infinity as x goes to infinity along any ray y = lambda x.

Now we examine the behaviour of f(x,y) on y=-1, here:

f(x) = f(x,-1) = 2 x^4 +x^2 -1

f'(x) = 8x^3 + 2x

which is 0 when x=0, which is a minimum where f(x,y) = -1

Now check the points y=-1 x=+/- 2, where f(x,y)=35

Now lets look at f(x,y) on the curve y=3-x^2.
Substiture this into the expression for f(x,y) to get, on this curve:

f(x) = -x^6 +12 x^4 -30 x^2 +27

and so f'(x)=0 has solutions x=0, x=+/-sqrt(4-sqrt(11)) abd x=+/-sqrt(4+sqrt(11)).
The last two roots (x=+/-sqrt(4+sqrt(11))) are outside our feasible region and so
are of no interest.

f(0)=27, f(+/-sqrt(4-sqrt(11))) ~= 11.78

So examining all these points we see that the minimum is acheived at (0,-1) where
it is -1, and the maximum is at (+/-2, -1) where it is 35.

RonL