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**careless25****Real Member**- Registered: 2008-07-24
- Posts: 560

A circular hot plate given by the relationship x^2 + y^2 <= 4 is heated according to the spatial temperature function T(x, y) = 10-x^2 + 2x - 4y^2. Find the hottest and coldest temperatures on the plate and the points at which they occur.

Here is what I have so far:

g(x,y) = x^2 + y^2 - K where K<= 4

del f = (-2x + 2, -8y)

del g = (2x, 2y)

Applying LaGrange multiplier

1) -2x+2 = lambda(2x)

2) -8y = lambda(2y)

3) x^2 + y^2 = K

solving for eqn 2)

we get y=0 or lamba = -4

sub lambda = -4 in eqn 1

we get x = -1/3

now I am stuck, since there are infinite values of K we have to solve for. I know there is going to be points (-1/3, +/- y) and (+/-x, 0).

Another approach that gives me half the answer is solving for the global max of T(x,y), which coincidentally lies in the constraint given, but then I am at a loss at finding the local min.

Any ideas?

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,278

hi careless25

I get Tmax = 11 at (1,0)

There isn't a global min within the range, in fact there won't be one for any limited range of x and y because T decreases as x and y get larger, indefinitely.

So I argued like this:

(i) Pick the value of y that makes T as small as possible. y=2

(ii) Use that for T . T = -x^2 + 2x -6

(iii) Find the lowest T. As this quadratic gets smaller as x gets larger that will be when x = 0. (0,2) gives Tmin = -6

I've checked this out in Excel using a range of values and confirmed both answers numerically.

LATER EDIT AFTER READING POST #4

I seem to have forgotten negative values so this is NOT the minimum.

Bob

ps. Thinking about it some more, I think this only works for these functions of x and y. You could probably construct a quadratic in x that doesn't have it's minimum when y has its minimum (by making the x^2 term positive for instance) so this approach wouldn't always work.

Children are not defined by school ...........The Fonz

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

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi careless25;

I have never seen a Lagrangian Multiplier used on an inequality. Usually the constraints and the function to be minimized/maximized are equations.

1) -2x+2 = lambda(2x)

2) -8y = lambda(2y)

3) x^2 + y^2 = K

Mathematica can take us to a complete solution to this but I do not understand it.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**TheDude****Member**- Registered: 2007-10-23
- Posts: 361

I think this is right.

First take the partial derivatives of T(x, y) and set them equal to 0:

Which agrees with Bob's answer. This is the only critical point for T(x, y) in the region x^2 + y^2 <= 4 (indeed, it's the only critical point anywhere). This means that to find the minimum of T(x, y) in this region we only need to consider the boundary x^2 + y^2 = 4.

Find the critical point:

Now we have just 3 points to check by hand, the boundary cases (-2, 0) and (2, 0) and the critical point (-1/3, +- sqrt{35/9}):

So the minimum temperature is -57/9 and occurs at (-1/3, +- sqrt{35/9})

Wrap it in bacon

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,278

Thanks The Dude,

For some reason I completely forgot about negative values of x. Your answer looks much better to me.

Bob

Children are not defined by school ...........The Fonz

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

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**careless25****Real Member**- Registered: 2008-07-24
- Posts: 560

TheDude,

How do we know that we just have to consider the boundary points? What if there was a graph whos min was not on the boundary point?

Thanks

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**TheDude****Member**- Registered: 2007-10-23
- Posts: 361

If it wasn't on a boundary point then it would be at a critical point, and we already know that there is only 1 critical point for T(x, y) and that it is a global maximum.

Wrap it in bacon

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