lagrange multipliermultivariable-calculus
I have the function : $$f(x,y,z)= x+y^2+3z^2$$ and I am asked to find the minimum and maximum of it in the cylinder $x^2+y^2 \leq4$.
firstly I dont understand why $x^2+y^2 \leq4$ is called cylinder and not circle, secondly how I can build my lagrange function with constraint that is expressed in inequality
A possible source of confusion here as to the use of the Lagrange multiplier $ \ \lambda \ $ appears to be due to the first Lagrange equation being factored incorrectly. The two equations are found properly, but we should have
$$ 3x^2 \ - \ 2 \lambda \ x \ = \ 0 \ \ \Rightarrow \ \ x \ ( \ 3x \ - \ 2 \lambda \ ) \ = \ 0 \ \ , $$
$$ 8y \ - \ 2 \lambda \ y \ = \ 0 \ \ \Rightarrow \ \ 2y \ ( \ 4 \ - \ \lambda \ ) \ = \ 0 \ \ . $$
So two of the possible solution sets are correctly given as $ \ ( 0 , \ \pm 1 ) \ $ and $ \ ( \pm 1, \ 0) \ $ by applying the constraint equation.
However, there is only the only value for the Lagrange multiplier found from the second equation, $ \ \lambda \ = \ 4 \ $ . This would be applied in the second factor of the first equation to obtain
$$ 3x \ - \ 2 \lambda \ = \ 0 \ \ \Rightarrow \ \ x \ = \ \frac{2 \cdot 4}{3} \ = \ \frac{8}{3} \ \ . $$
This raises an interesting issue in problems of this sort. In this case, the result produced by using the Lagrange multiplier value is inapplicable because there are no points with $ \ x \ = \ \frac{8}{3} \ $ on the constraint surface. [A similar circumstance arises in this problem.] Therefore, the extremal values of our function are
$$ f(0, \ \pm 1) \ = \ 4 \ \ [\text{absolute maximum}] \ \ ; $$
$$ f(1, \ 0) \ = \ 1 \ \ [\text{local maximum}] \ \ ; $$
$$ f(-1, \ 0) \ = \ -1 \ \ [\text{absolute minimum}] \ \ . $$
The "geometric" interpretation of this result is that the values for the function represent maximum and minimum values of the $ \ z-$ coordinate of the possible intersection points of the surface $ \ z \ = \ f(x,y) \ = \ x^3 \ + \ 4y^2 \ $ with the "vertical" cylinder $ \ x^2 \ + \ y^2 \ = \ 1 \ $ .
We show two views of the intersecting surfaces, with three of the four "critical points" visible; $ \ (0, \ -1, \ 4) \ $ is on the "far side" of the diagram at left.
$$ \ \ $$
The value found for the multiplier is the same regardless of the radius of that cylinder, which suggests that the coordinate $ \ x \ = \ \frac{8}{3} \ $ could become important if the constraint surface were "larger". If we were to use the cylinder $ \ x^2 \ + \ y^2 \ = \ 9 \ $ as the constraint "surface", for instance, our result from the Lagrange multiplier would be applicable, now giving us extremal values
$$ f(0, \ \pm 3) \ = \ 36 \ \ [\text{absolute maximum}] \ \ ; $$
$$ f(3, \ 0) \ = \ 27 \ \ [\text{local maximum}] \ \ ; $$
$$ f(-3, \ 0) \ = \ -27 \ \ [\text{absolute minimum}] \ \ ; $$
$$ f\left(\frac{8}{3}, \ \pm \frac{\sqrt{17}}{3}\right) \ = \ \frac{716}{27} \ \approx \ 26.52 \ \ [\text{local minima}] \ \ . $$
For these larger cylindrical radii, the intersection with the surface $ \ f(x,y) \ $ along the "positive $ -x \ $ " face of the cylinder develops a "ripple", leading to the appearance of further "critical points" with $ \ x \ = \ \frac{8}{3} \ $ .
The somewhat more complicated shape of the intersection of the surfaces is visible in the left-hand graph; the critical point $ \ (0, \ -3, \ 36) \ $ is not visible in these views.
It is a closed region, so max and min must occur. They can only occur on the boundary or at critical points of the function. So you can use the following steps:
Step 1: Find all the critical points of the function, and check whether they are in the constraint region.
Step 2: Use regular Lagrange multiplier method on the boundary of the disk.
Then combine the results from the two steps to find the max and min.
Best Answer
$x^2+y^2 \leq4$ is a cylinder not a circle because you are working in the three dimensional plane plane, see how your main function is in terms of $x,y,z$. This would be a circle going all along the z-axis (you can imagine in as a non-ending infinite can), a cylinder in three dimensions always has $2$ variables, and it goes along the third missing one. To draw cylinders in three dimensions draw the function given (the circle in your case, could be any curve in general) then extend it so it goes all along the missing variable.
Concerning the set-up, use the following:
$g(x,y,z)=x^2+y^2-4\leq0$ (constraint)
$f(x,y,z)=x+y^2+3z^2$ (function)
We set-up and solve the following system of equations:
$\nabla\vec f=\lambda\nabla\vec g$ ($\nabla$ means the gradient)
$g(x,y,z)=0$
$\nabla\vec f=1i\vec +2y\vec j+6z\vec k$
$\nabla\vec g=2x\vec i+2y\vec j$
so we get the following system:
$2x=\lambda$ $(1)$
$2y=\lambda 2y$ $(2)$
$0=\lambda0=0$ $(3)$
$x^2+y^2-4 \leq 4$ $(4)$
You might get multiple coordinates, all coordinates you will get will represent the points where there is an extreme value. To check which gives you a maximum and which gives you a minimum you should plug them in $f(x,y,z)$ and see compare the values. Pay attention that you might get points which aren't on the cylinder, in this case you should rule them out.
Additional tip based on personal experience:
Usually the the challenge in Lagrange problems is to identify the function and constraint. Example of keywords that might help in identifying the constraint: "lying on the surface", "can be inscribed in", "surface area equal to S(some number)", etc.... Of course other words can be used and you should always think about what you choose as a constraint, but these might be good as a hint.