Easy (?) application of Lagrange multiplier

Question

I am reading a book about utility theory and there is a exercise (without solution). I can't stop thinking about this, since the normal Lagrange multiplier approach seems not to work.

We want to maximize $U(x,y,z)=xy+az$ where $a>0$ under the restriction of $x+y+z=100$.

When applying the Lagrange method we get the Lagrange function $$\Lambda(x,y,z,\lambda)=xy+az+\lambda(100-x-y-z)$$ When we take the partial derivatives we get $$\frac{d\Lambda}{d\lambda}=100-x-y-z$$ $$\frac{d\Lambda}{dx}=y-\lambda$$$$\frac{d\Lambda}{dy}=x-\lambda$$$$\frac{d\Lambda}{dz}=a-\lambda$$ So we can derive that $y=x=a$ and $z=100-2a$ in the optimum. Besides domain issues one can find easy counterexamples like a=10. For $y=x=10$ and $z=80$ we get $$U(10,10,80)=900<2500=U(50,50,0)$$ Why does the Lagrange method fail here? (Or have I failed using the method?)

Answer 1

This is too long for comment. Here the utility function is Cobb-Douglas in $x$ and $y$. The optimization agent would consume equal amounts of $x$ and $y$. Given his budget set, the highest marginal utilities of $x$ and $y$, he can achieve is $50$. The marginal utility of $z$ on the other hand is the constant $a$. The agent is trading off the marginal utility he gets from $z$ and the two complimentary goods $x$ and $y$.

So the optimal bundle should depend on the value of $a$. There exist threshold $a_L$ such that

1. If $a < a_L$, i.e. when $a$ is sufficiently low, then no $z$ is consumed. Optimal bundle is $(50, 50, 0)$. For this to happen,

$$ 100 a \leq 2500 \Rightarrow a_L = 25. $$

2. If $a_L \leq a$, then optimal bundle is $(0,0,100)$.

The interior/Lagrange multiplier solution is never optimal since $100a \geq 100a - a^2$.