Lagrange Multiplier inconsistent

Question

max $f(x,y,z)$ subject to $x+y+z=1$

$$f(x,y,z)= - (x+0.5y)\log(x+0.5y) -2(0.25y+0.5z)\log(0.25y+0.5z)-\log(2)(1.5y+z)$$

$$\frac{\partial F}{\partial x} = - (1+\log(x+0.5y))$$ $$\frac{\partial F}{\partial y} = -0.5(1+\log(x+0.5y)) - 0.5(1+\log(0.25y+0.5z) -1.5\log(2)$$ $$\frac{\partial F}{\partial z} = - (1+\log(0.25y+0.5z)-\log(2)$$

setting $$\frac{\partial F}{\partial x} = \lambda$$ $$\frac{\partial F}{\partial y} = \lambda$$ $$\frac{\partial F}{\partial z} = \lambda$$ $$x+y+z=1$$

and solving with $$\frac{\partial F}{\partial y} =\frac{\partial F}{\partial x} $$ we get $$x+0.5y=e^{3\log(2)}\left(0.25y+0.5z\right).$$

similarly solving with $$\frac{\partial F}{\partial x} =\frac{\partial F}{\partial z} $$ we get $$x+0.5y=e^{\log(2)}\left(0.25y+0.5z\right).$$

The two equations obtained by solving for $\lambda$ are inconsistent. Why ?

Answer 1

I think there exists no solution, which would also explain the inconsistency. I found this by transforming your problem to apropriate coordinates

$$X=x+\frac{y}{2} \\ Y=\frac{y}{4}+\frac{z}{2}\\ Z=\frac{3y}{2}+z$$

The side conditions then become $X+2Y=0$ resulting in the Lagrange function

$$L(X,Y,Z,\Lambda)=F(X,Y,Z)+\Lambda(X+2Y)$$

with $$F(X,Y,Z)=-X\log X -2Y\log Y -Z\log2.$$

Your equations are

$$0 =L_\Lambda=X+2Y\\ 0 =L_X=-\log X-1+\Lambda\\ 0 =L_Y=-2\log Y-2+2\Lambda\\ 0 =L_Z=-\log 2$$

where the last equation cannot be solved. If I didn't make a mistake there is no extremum.

Answer 2

I am assuming that we are working with natural logarithms (this is not essential to the outcome -- and I won't ask where you acquired such a lovely function). I agree with your partial derivatives, so we have the results

$$\frac{\partial F}{\partial x} \ = \ \frac{\partial F}{\partial y} \ \ \Rightarrow \ \ x \ + \ \frac{1}{2}y \ = \ 8 \ \left(\frac{1}{4}y \ + \ \frac{1}{2}z\right) \ \ \Rightarrow \ \ x \ - \ \frac{3}{2} y \ - \ 4z \ = \ 0 \ \ , $$

$$\frac{\partial F}{\partial x} \ = \ \frac{\partial F}{\partial z} \ \ \Rightarrow \ \ x \ + \ \frac{1}{2}y \ = \ 2 \ \left(\frac{1}{4}y \ + \ \frac{1}{2}z\right) \ \ \Rightarrow \ \ x \ - \ z \ = \ 0 \ \ , $$

$$\frac{\partial F}{\partial y} \ = \ \frac{\partial F}{\partial z} \ \ \Rightarrow \ \ 2 \ \left(x \ + \ \frac{1}{2}y\right) \ = \ \frac{1}{4}y \ + \ \frac{1}{2}z \ \ \Rightarrow \ \ 2x \ + \ \frac{3}{4} y \ - \ \frac{1}{2}z \ = \ 0 \ \ . $$

Using whatever algebraic method you prefer, we arrive at $ \ x \ = \ -\frac{1}{2}y \ = \ z \ $ . The problem isn't that the system is inconsistent -- it is linearly dependent , and not in a good way. We find that $ \ x \ + \ \frac{1}{2}y \ = \ 0 \ $ , which is then the value of the argument of the logarithmic function in the first term of $ \ f(x,y,z) \ $ . Thus, this solution plane for candidate extremal points is not in the domain of the function.

The Lagrange-multiplier method seeks locations at which level surfaces of the function to be extremized are tangent to the surface(s) described by the constraint equation(s). This does not come with any promise that there are such points. An inconsistency in the system of Lagrange equations (often) means that there are no tangent points, either because there is no level surface that makes contact with the constraint surface(s) at all, or that the level surfaces can only cross the constraint surface(s). In this problem, the latter appears to be the case.

Here is a graph of the constraint plane, $ \ x \ + \ y \ + \ z \ = \ 1 \ $ , along with what would be the "planes of limit points" for the domain, $ \ x \ + \ \frac{1}{2}y \ = \ 0 \ $ and $ \ \frac{1}{4}y \ + \ \frac{1}{2}z \ = \ 0 \ $ . The magneta strip marks the portion of the constraint plane which lie in the domain of $ \ f(x,y,z) \ $ .

Unfortunately, it isn't possible to show a graph of the function itself, as it has three independent variables. The diagram below depicts three-dimensional "cross-sections" representing $ \ f(x,y,0) \ , f(x,y,1) \ , f(x,y,2) \ $ . (Negative values for $ \ z \ $ produce surfaces that don't even reach the constraint plane.) It appears that the function "arches over" or crosses through the available portion of the constraint plane, without having tangent points. So the function does not seem to have extrema subject to the constraint.