Finding max/min given contraint

[ultra100]

Homework Statement

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

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?

 

[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.

 

[ultra100]

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

 

[NoMoreExams]

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

Refresh your page :)

 

[ultra100]

Thanks! This helps a lot