Here's the function: $f(x,y,z)=xyz^3$ under the constraint $3x^2+2y^2+z^2=1$
My effort:
$yz^3=\lambda6x$
$xz^3=\lambda4y$
$3xyz^2=\lambda2z$
But from there I don't know how to solve this system of equation.
Any help to point me in the right direction would be greatly appreciated.
Thank you.
On
The point of the Lagrange multiplier strategy is that the gradient vector of the function $g(x,y,z) = 3x^2 + 2y^2 + z^2$ is perpendicular to the surface defined by g(x,y,z) = 1, since the function is constant on that surface. If the function $f(x,y,z) = xyz^3$ has a maximum or minimum value on this surface, then the gradient at that point must also be perpendicular to the surface.
If it was not perpendicular, that means we could move in the direction of the gradient along the surface and get to either higher or lower values, and in the opposite direction to get the opposite type of values, contradicting the idea that this point was a maximum or minimum. Since both gradient vectors are perpendicular to the surface at the same point, and thus parallel to each other, we may search for this occurrence by looking for a point where the gradient vectors are a scalar multiple, $\lambda$, of each other.
From the strategy, we must first find the gradient vectors of each function. $$\nabla f(x,y,z) = \begin{pmatrix}\frac{\partial f}{\partial x}\\ \frac{\partial f}{\partial y}\\ \frac{\partial f}{\partial z}\end{pmatrix} = \begin{pmatrix}yz^3\\ xz^3\\ 3xyz^2\end{pmatrix}$$ $$\nabla g(x,y,z) = \begin{pmatrix}\frac{\partial g}{\partial x}\\ \frac{\partial g}{\partial y}\\ \frac{\partial g}{\partial z}\end{pmatrix} = \begin{pmatrix}6x\\ 4y\\ 2z\end{pmatrix}$$ Now we get three equations by setting the gradient vector of f equal to an unknown scalar multiple, $\lambda$, of the gradient vector of g. Its value is only used to solve for x, y and z. It is otherwise unimportant: $$yz^3 = \lambda\cdot 6x$$ $$xz^3 = \lambda\cdot 4y$$ $$3xyz^2 = \lambda\cdot 2z$$ Because we have 4 unknown variables, x, y, z, and $\lambda$, we will need the fourth equation given by the constraint, $3x^2 + 2y^2 + z^2 = 1$. You can get the same system of 4 equations by taking the derivative of the function $f - \lambda g$, which has a similar geometric interpretation.
Since $z^3$ is easy to isolate in the first two equations, assuming $x\neq 0$ and $y\neq 0$, we may equate the two values of $z^3$. We will explore the two excluded cases later. $$6\lambda\frac{x}{y} = 4\lambda\frac{y}{x}$$ We may solve this to get x in terms of y: $$x = \pm y\sqrt{\frac{2}{3}}$$ Using this value, we may solve for z in terms of y by substitution of this value for x in the constraint equation: $$z = \pm\sqrt{1 - 4y^2}$$ Finally, we may use substitution of these two values of x and z in the third component equation to solve for $\lambda$ in terms of y. Assuming $z\neq 0$, we may divide both sides of the third component equation by z to get: $$\lambda = \pm \sqrt{\frac{3}{2}}y^2\sqrt{1 - 4y^2}$$ Now we may use any equation to attempt to solve for y, substituting in our values for x, z, and $\lambda$. Let us use the first component equation: $$y(1 - 4y^2)^\frac{3}{2} = \pm 6y^3\sqrt{1 - 4y^2}$$ One solution is y = 0, a case we avoided earlier. If y = 0, the remaining 3 equations become: $$xz^3 = 0$$ $$0 = 2\lambda z$$ $$z^2 = 1$$ This implies $z = \pm 1$ and x = 0. So two potential extrema are $(0, 0, \pm 1)$. If $y\neq 0$, we may divide both sides by y to obtain: $$(1 - 4y^2)^\frac{3}{2} = \pm 6y^2\sqrt{1 - 4y^2}$$ Two solutions of this equation are $y = \pm\frac{1}{2}$. If y is this value, the remaining three equations become: $$xz^3 = \pm 2\lambda$$ $$\pm\frac{3}{2}xz^2 = 2\lambda z$$ $$1 + z^2 = 1$$ This implies that z = 0 and $x = \pm\frac{1}{\sqrt{6}}$, where the latter value comes from using the first two component equations to derive the relationship $6x^2 = 4y^2$ when neither x nor y are 0. So 4 more solutions of the system of equations are $\left(\pm\frac{1}{\sqrt{6}}, \pm\frac{1}{2}, 0\right)$. If $1 - 4y^2\neq 0$, then we may divide both sides of the equation by $\sqrt{1 - 4y^2}$ to get: $$1 - 4y^2 = \pm 6y^2$$ which has the solutions $y = \pm\frac{1}{\sqrt{10}}$. Using these values in the remaining three equations, we get 6 solutions: $\left(\pm\frac{1}{\sqrt{15}}, \pm\frac{1}{\sqrt{10}}, \pm\sqrt{\frac{3}{5}}\right)$. Now, from our study of types of critical points, it is possible that these points are not extrema, but are just stationary points of some other type, such as saddle points. So to complete the problem, you must use the second derivative of $f - \lambda g$, evaluated at each of the 12 points. It is a matrix representing the map from displacement vectors of gradient vectors at each point to the linear change in the gradient vector in that direction. We therefore want to check to see if the matrix is positive or negative definite, which implies all directions from the point are associated with gradient vectors changing in the same direction. Indefinite or singular matrices do not represent extrema, as they represent either directions in which the gradient does not change, or gradients in some directions have an inverted direction to gradients changing in another direction.
Edit 2: One final edit to say that I recommend looking at Ronald Lett's answer, as he did what I was trying not to do and addressed all possible cases. He even did so in a thorough and, I believe understandable, answer.
You need to define a lagrangrian, $\mathscr{L}$, which is your objective function (the function you are minimizing/maximizing) minus a constant (the lagrange multiplier), we can call this $\lambda$ times your constant (set $=0$, so in your case you need to subtract the $1$ from both sides).
So for your problem we have $\mathscr{L} = xyz^3 - \lambda (3x^2+2y^2+z^2 -1)$
Now you want to take the derivative of $\mathscr{L}$ with respect to whatever you are minimizing against (you don't specify in your problem? I'm assuming $x,y$, and $z$), set these derivatives equal to 0, and solve. Call the resulting $\lambda, x,y,z$ you get $x^*,y^*, z^*$, which would be your solutions.
This is a sketchy explanation, but you should help your progress, and hopefully either someone else provides fuller answer or perhaps I shall edit this later.
Edit: I answered the wrong thing.
This may be wrong as well , but what about rearranging so the right side of all 3 equations you have is $2\lambda$ then subtract the first from the second and get $2y^2 = 3x^2$. Then set the first and third equal and get $9x^2 = z^2$. Then plug these into your constraint to find z, then use z to get x and x to get y.
I believe this is a solution, but perhaps not the minimum.