Finding max/min given contraint

  • Thread starter ultra100
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In summary, the maximum and minimum values of the function f(x,y) = xy on the ellipse (x^2)(1/9) + y^2 = 2 are x = (-1,3), y = (-1,3). The Attempt at a Solution states that I tried using Lagrange Multipliers and got y = \pm 1 .
  • #1
ultra100
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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?
 
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  • #2
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.
 
Last edited:
  • #3
what equations are you using to get y = to +/- 1?
 
  • #4
ultra100 said:
what equations are you using to get y = to +/- 1?

Refresh your page :)
 
  • #5
Thanks! This helps a lot
 

1. What is the definition of a constraint in the context of finding max/min?

A constraint is a limitation or restriction that must be satisfied when finding the maximum or minimum value of a function. It can be in the form of an equation, inequality, or condition that the function must adhere to.

2. How do you identify constraints when finding max/min?

To identify constraints, you first need to determine the independent and dependent variables in the function. Then, look for any conditions or restrictions that these variables must satisfy. These conditions will serve as the constraints for finding the maximum or minimum value of the function.

3. What is the role of Lagrange multipliers in finding max/min given constraints?

Lagrange multipliers are used to optimize a function subject to constraints. They help in finding the maximum or minimum value of a function while satisfying the given constraints. This method involves finding the critical points of the function and checking for any potential maxima or minima.

4. How do you solve for max/min given constraints using the method of Lagrange multipliers?

To solve for max/min using Lagrange multipliers, first set up the Lagrangian function by adding the product of the constraints and the Lagrange multiplier to the original function. Then, find the partial derivatives of the Lagrangian function and set them equal to zero. Solve the resulting system of equations to find the values of the variables that satisfy both the function and the constraints.

5. Can the method of Lagrange multipliers be used for all types of constraints?

Yes, the method of Lagrange multipliers can be used for both equality and inequality constraints. For equality constraints, the constraint equation is set equal to the Lagrange multiplier, while for inequality constraints, an additional step of checking for feasibility is required after solving the system of equations.

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