# Maxima and Minima on surfaces in three dimensional space

• theno1katzman
In summary: So differentiate and solve for df/dx = 0. You get 4x(1-x2) = 0, x \in [0, 1]. Now if you can't see from the start that this is where the max and min are going to be, try points a bit more than 1 and a bit less than 1, and if you don't see what's happening, try plotting the graph. You will see that x = 0 and x = 1 are the only critical points in the domain and that the values of the function at these two points are 0 and 1. So

## Homework Statement

Find the maximum and minimum values of f(x,y) = (xy)2 on the domain x2 + y2 < 2. Be sure to indicate which is which

## Homework Equations

I am not sure what to put here. I solved this problem a different way, and I am not confident I did it correctly.

## The Attempt at a Solution

fx = 2xy2 = 0
fy = 2x2y = 0

only critical point is (0,0) and the value of f there is f(0,0)= 0. Therefore the local and absolute minimum value is 0.

I tested to make sure by estimating what the graph would look like, I have a lot of trouble graphing the traces once I do this on the three coordinate axes. It has hyperbolas in the first and third quadrants as xy traces. It has parabolas becoming ever steeper for the yz and xz traces with the vertex of them all being at (0,0) which proves that that the minimim is at the origin. I never took the restricted domain into account however because I thought it was irrelevant. Which makes me think I did something wrong. The restricted domain is a circle of radius 2.

Hope you can understand what I did, and thanks for your help!

Maxima/minima don't have to occur just at critical points; they can also occur at endpoints of the domain (for functions of a single variable) or at boundaries of domains (for functions of more than one variable).

You need to look at the boundary of the domain.

Indeed you have to look at the boundary.

and your equations give more solutions than you think,
x=0 for any y
y=0 for any x

The radius for the domain boundary is actually [sqrt]2.

Some more work I tried using this method the textbook showed me, however the textbook was easier and had a different equation for the surface.

4 points to try on the domain where one of the values of the function is zero are
(0,sqrt 2) (0,-sqrt 2) (sqrt 2,0) (-sqrt 2,0)

I am not sure what to say, here, the minimum vales of those are all going to be 0.

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I don't know whether I have misunderstood this problem, but isn't (xy)2 = K four families of hyperbolae? By symmetry aren't the ones with the largest K is the ones going through

$$(\sqrt{2},\sqrt{2})$$ , $$(-\sqrt{2},\sqrt{2})$$, $$(\sqrt{2},-\sqrt{2})$$ and $$(-\sqrt{2},-\sqrt{2})$$ ? for which in each case (xy)2 = $$((\sqrt{2})^2) = 4$$. Minimum values for (xy)2 are 0. Obviously it is never going to be less than that.

Then for formal machinery to get the result this is an exercise in constrained maximum.

epenquin, the points you mention are actually outside the the domain...
but indeed with some insight in the function you easily see the solutions.

A more general approach to finding min/maxima at the domain edge for example to parametrize the boundary.
so take
x= sqrt 2 cos(t)
y= sqrt 2 sin(t)

now f(t)=(2 cos(t) sin(t) )^2 = sin(2t)^2,
setting the derivative to zero you find solutions: t=k *pi/4 for k=0...7.
and there you have your max and min on the boundary of the domain.

I still doubt you understand my previous post thought, it actually states that on the x- and y-axis the function is at a minimum. so your 4 points are actually trivial since you already know them.

justsof said:
epenquin, the points you mention are actually outside the the domain...
but indeed with some insight in the function you easily see the solutions.
Oops, silly slip, I should have said max of (xy)2 on the domain is 1 - at the points (1, 1) (1, -1) (-1,1) and (-1, -1). Still same idea though.

justsof said:
I still doubt you understand my previous post thought, it actually states that on the x- and y-axis the function is at a minimum. so your 4 points are actually trivial since you already know them.
Yes I agree the minimum value, 0, of the function is realized at every point on the x, y axes.

min value would be 0 and max value would be 1

Firstly I do not see that this is a problem of 3 dimensions, for me it is 2. But a similar problem can very well arise in 3 dimensions, or any number. As a matter of fact almost the same problem in 3 dimensions, max/minimising xyz on a sphere did arise for me in a biophysical problem.

Important to realize this is really two different problems. Your solving fx = 0, fy = 0, gives you all the extrema in the plane as a whole. Then you see whether any of these are within the disk f<$$\sqrt{2}$$ and in this case we find a whole pair of lines where they are. From these, four points are actually on the circle x2 + y2 = 2.

But then a different question is to find any points that are not extrema in the plane as a whole but are the max or min values of f encountered as you go around the circle. For any continuous closed curve unless f is simply constant there you are bound to have at least one max and one min of the function which in general are not extrema in the plane as a whole. But the problem need not be about closed curves - you might be (soon will be) asked to find the max/min of some f(x, y) along a curve g(x, y) = 0 between x = a and x = b for instance. (In such a case x=a and x=b are candidates).

In simple cases you can solve it by eliminating a variable. In this case you could express the equation in polar co-ordinates and as the circle is r= constant you reduce it to finding where ($$cos\theta.sin\theta)^2$$ has max and I think you'll get a nice thing involving 4$$\theta$$ by standard manips. I am a bit too error-prone to give you that, but I found it very easy keeping the original x,y co-ordinates. You have along the circle f = 2x2 - x4 and a similar equation for y, which you can now differentiate (this is now now df/dx as you go along the circle) you easily find solutions $$x = \pm1, y=\pm1$$.

You can see that in less simple cases such an elimination may not be possible or convenient and then you have to use another method such as Lagrange's or equivalent which you are probably being introduced to.

Well I'm not sure I understand everything you wrote there, I think the problem was simpler when converted to polar coordinates.
Thanks for your help and suggestion to convert to polar coordinates.

Here is the problem worked out once more.

t is theta when converted to polar coordinates, not a parametric.
f(x,y) =(xy)2 on domain
x2+y2<2

fx=2(xy)y=2xy2 => fr=2rcostr2sin2t=0

fy=2(xy)x=2x2y => ft=2rcos2trsint=0

Since they both equal zero, set them equal to each other.

2r3costsin2t = 2r3cos2tsint

this simplify to sint=cost and the only values for t that makes sense are pi/4 or 5pi/4

placing the values back into the original equation in polar coordinates
f(r,t)=(r2sintcost)2=0
yields r being equal to zero.

so critical points in the domain of r2<2 will be
f(0,pi/4)=0 or f(0,5pi/4)=0
f(r,pi/4)=(r2/2)2 for domain 0<r2<2
minimum value here is 0
maximum value here is 1

f(r,5pi/4)=(r2/2)2 for domain 0<r2<2
minimum value here is 0
maximum value here is 1

The maximum value of the function is 1. The minimum value of the function is 0.