Finding max/min given contraint


by ultra100
Tags: constraint equation, lagrange, maximum, minimum, multivariable
ultra100
ultra100 is offline
#1
Dec15-08, 09:13 PM
P: 9
1. The problem statement, all variables and given/known data

Find the product of the maximum and minimum values of the function f(x, y) = xy
on the ellipse (x^2)(1/9) + y^2 = 2


3. The attempt at a solution

I tried soving using lagrange multiplier and got:

fx = y - (2/9)(x*lambda)
fy = x - 2y*lambda
flambda = (x^2)(1/9) + y^2 - 2

then I set these = 0, but my answer came out wrong.. any suggestions for figuring out the min/max?
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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NoMoreExams
NoMoreExams is offline
#2
Dec15-08, 09:23 PM
P: 626
For y I'm getting, [tex] y = \pm 1 [/tex]

The way I set this up was as follows:

Define:

[tex] f(x,y) = xy; g(x,y) = \frac{x^{2}}{9} + y^{2} = 2 \Rightarrow h(x,y,\lambda) = f(x,y) + \lambda g(x,y) [/tex]

(Note that the sign in front of [tex] \lambda [/tex] does not matter)

So let's take our partials, we get:

[tex] \frac{dh}{dx} = y + \frac{2 \lambda}{9}x, \frac{dh}{dy} = x + 2 \lambda y, \frac{dh}{d\lambda} = \frac{x^2}{9} +y^2 - 2 [/tex]

We know that each of those partials vanish i.e. we can set each to 0.

The first one gives us

[tex] 9y = -2 \lambda x [/tex]

and the second one gives us

[tex] \frac{x}{y} = -2 \lambda[/tex]

And by simple substitution we get:

[tex] 9y^{2} = x^{2} [/tex]

So let's substitute it into our 3rd equation to get:

[tex] y^{2} + y^{2} = 2 [/tex]

Which yields our desired result of [tex] y = \pm 1 [/tex]. Now we can plug this into our g(x,y) to get [tex] x = \pm 3 [/tex]

Note that it doesn't matter which value for y we pick therefore our solution set will be:

[tex] (1,3), (1,-3), (-1,3), (-1,-3) [/tex]

Now if you don't want to do this using Lagrange Multipliers, we can just realize that we can rewrite our g(x,y) as

[tex] y = \pm \sqrt{2 - \frac{x^2}{9}} [/tex]

and now we can substitute this into our f(x,y) get an equation of one variable i.e.

[tex] \bar{f}(x) = \pm x \cdot \sqrt{2 - \frac{x^2}{9}} [/tex]

Now we can proceed using the techniques you learned in Calculus 1 (I'm going to use Maple because I'm lazy)

> a:=x*sqrt(2-x^2/9);

a := [tex] \frac{1}{3} x \sqrt{18 - x}[/tex]

> solve(diff(a,x)=0,x);

-3, 3

Note that choosing the negative root produces the same results.
ultra100
ultra100 is offline
#3
Dec15-08, 09:31 PM
P: 9
what equations are you using to get y = to +/- 1?

NoMoreExams
NoMoreExams is offline
#4
Dec15-08, 09:50 PM
P: 626

Finding max/min given contraint


Quote Quote by ultra100 View Post
what equations are you using to get y = to +/- 1?
Refresh your page :)
ultra100
ultra100 is offline
#5
Dec15-08, 09:56 PM
P: 9
Thanks!!! This helps a lot


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