# Homework Help: Finding max/min given contraint

1. Dec 15, 2008

### ultra100

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. Dec 15, 2008

### NoMoreExams

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.

Last edited: Dec 15, 2008
3. Dec 15, 2008

### ultra100

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

4. Dec 15, 2008