# Finding max/min given contraint

 P: 626 For y I'm getting, $$y = \pm 1$$ The way I set this up was as follows: Define: $$f(x,y) = xy; g(x,y) = \frac{x^{2}}{9} + y^{2} = 2 \Rightarrow h(x,y,\lambda) = f(x,y) + \lambda g(x,y)$$ (Note that the sign in front of $$\lambda$$ does not matter) So let's take our partials, we get: $$\frac{dh}{dx} = y + \frac{2 \lambda}{9}x, \frac{dh}{dy} = x + 2 \lambda y, \frac{dh}{d\lambda} = \frac{x^2}{9} +y^2 - 2$$ We know that each of those partials vanish i.e. we can set each to 0. The first one gives us $$9y = -2 \lambda x$$ and the second one gives us $$\frac{x}{y} = -2 \lambda$$ And by simple substitution we get: $$9y^{2} = x^{2}$$ So let's substitute it into our 3rd equation to get: $$y^{2} + y^{2} = 2$$ Which yields our desired result of $$y = \pm 1$$. Now we can plug this into our g(x,y) to get $$x = \pm 3$$ Note that it doesn't matter which value for y we pick therefore our solution set will be: $$(1,3), (1,-3), (-1,3), (-1,-3)$$ 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 $$y = \pm \sqrt{2 - \frac{x^2}{9}}$$ and now we can substitute this into our f(x,y) get an equation of one variable i.e. $$\bar{f}(x) = \pm x \cdot \sqrt{2 - \frac{x^2}{9}}$$ 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 := $$\frac{1}{3} x \sqrt{18 - x}$$ > solve(diff(a,x)=0,x); -3, 3 Note that choosing the negative root produces the same results.