Finding max/min given contraintby ultra100 Tags: constraint equation, lagrange, maximum, minimum, multivariable 

#1
Dec1508, 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 



#2
Dec1508, 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(2x^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. 



#3
Dec1508, 09:31 PM

P: 9

what equations are you using to get y = to +/ 1?




#4
Dec1508, 09:50 PM

P: 626

Finding max/min given contraint 



#5
Dec1508, 09:56 PM

P: 9

Thanks!!! This helps a lot



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