2
$\begingroup$

I'm trying to find the max and mins of the equation $f(x,y,z) = xy + 3xz + 2yz$ on the constraint, $g(x,y,z)=5x+9y+z-10$. So according to the Lagrange Multiplier procedure, I take the partial derivatives of both equations and get,

$\frac{\partial f}{\partial x} = y + 3z$, $\frac{\partial g}{\partial x} = 5$

$\frac{\partial f}{\partial y} = x + 2z$, $\frac{\partial g}{\partial y} = 9$

$\frac{\partial f}{\partial z} = 3x + 2y$, $\frac{\partial g}{\partial z} = 1$

Then, using $\nabla f = \lambda\nabla g$

$y+3z = 5\lambda$

$x+2z = 9\lambda$

$3x+2y = \lambda$

However, the problem I encounter now is that I am having trouble getting the variables x, y, and z in terms of one variable when solving for lambda. If I am unable to do so, does this mean that the maxima and minima are undefined? I also noticed that the constraint was a plane, so perhaps there can't be any extrema because the constraint equation runs on values of x, y, and z that can take range from $\infty$ to $-\infty$ (no restriction on domain)?

$\endgroup$
3
  • $\begingroup$ Note that you also have another equation, $5x+9y+z=10$, meaning you have 4 equations in 4 variables. $\endgroup$ – πr8 Apr 5 '16 at 15:32
  • $\begingroup$ How should I utilize this 4th equation? Because I'm still having trouble getting the variables in terms of one variable. $\endgroup$ – Chris Gong Apr 5 '16 at 15:34
  • $\begingroup$ Well, for a start, $3(9\lambda)-\lambda=3(x+2z)-(3x+2y)=6z-2y \implies 3z-y=13\lambda$. So you should be able to find $y,z$ in terms of $\lambda$ from this and your other equation involving $y,z,\lambda$. $\endgroup$ – πr8 Apr 5 '16 at 15:38
2
$\begingroup$

One approach is to multiply the first equation by 9, the second by 5, the third by 45, giving all 3 equations a RHS of $45\lambda$, so we can set them equal to each other.

$9y+27z=45\lambda$

$5x+10z=45\lambda$

$135x+90y=45\lambda$

Set the first equal to the second, the second equal to the third, to give:

$9y+27z=5x+10z$

$5x+10z=135x+90y$

Now, simplify to the following:

$17z=5x-9y$

$10z=130x+90y$

Divide the second equation by 10 to get the pair of equations

$17z=5x-9y$

$z=13x+9y$

Now, try adding the equations, and substituting the result into one of the equations.

P.S. When approaching Lagrange Multipliers, I tend to avoid computing the value of $\lambda$ unless absolutely necessary, only resorting to this when literally everything else fails. One other thing: I have assumed $\lambda \neq 0$ throughout. Indeed, if $\lambda=0$, you can show that $x=y=z=0$ is the only solution to those 3 original equations, which doesn't satisfy the constraint.

$\endgroup$
4
  • $\begingroup$ Thank you for your help. Continuing from what you had, I was able to get a value for x,y, and z. x and z being -5/3 and y being 20/9 but the answer is that there are no extrema. I plugged x for z back into 9y + 27z and equated it to 135x + 90y. This got me y = -4/3x. Then I plugged the values for y and z back into the constraint to get it in terms of x to get y = 20/9 and x = -5/3. $\endgroup$ – Chris Gong Apr 5 '16 at 21:42
  • $\begingroup$ So I plugged the values I got back into the main equation. Then I tested other values that showed that the result was neither a max nor min so I guess it does not really have a max or min but I don't know why. Thanks for your help though. $\endgroup$ – Chris Gong Apr 6 '16 at 23:23
  • 1
    $\begingroup$ So with Lagrange Multipliers, talking about global extrema doesn't necessarily mean what you think it does. For instance, in this problem, you are trying to maximize or minimize $f(x,y,z)=xy+3xz+2yz$ on the plane $5x+9y+z=10$. This means that, while the function $f(x,y,z)$ has domain $\mathbb{R}^3$, and in fact has no global maxima or minima at all, we aren't concerned with the function on the whole of $\mathbb{R}^3$. $\endgroup$ – Nicholas Stull Apr 7 '16 at 15:53
  • 1
    $\begingroup$ Instead, we are concerned with the behavior of the function when we restrict its entire domain (its universe, so to speak) to the plane. In other words, the point you found might be a max or a min of the restriction of $f(x,y,z)$ to the plane, but it doesn't need to correspond to an extrema of $f(x,y,z)$ on the whole domain. $\endgroup$ – Nicholas Stull Apr 7 '16 at 15:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.