How to use Lagrange Multiplier for same degree?

Question

$$\begin{array}{ll} \text{minimize} & \overbrace{a_1 x^n + a_2 y^n + a_3 z^n}^{=: f(x,y,z)}\\ \text{subject to} & b_1x^n+b_2y^n+b_3z^n = k\end{array}$$

where not all $a_i = b_i$ for $i \in \{1,2,3\}$. Find the min/max value of function $f$ using the Lagrange multiplier method.

How can we do it? Applying partial derivative gives different values of $\lambda$ but no $\lambda$ satisfies every condition.

Any help is appreciated, Thanks.

Answer 1

The technique of lagrange multipliers has a geometric appeal which is the determination of tangency between surfaces. If we have the problem

$$ (\min|\max) f(x),\ \ \text{s.t.}\ \ g(x) = c $$

then after forming the lagrangian

$$ L(x,\lambda) = f(x)+\lambda(g(x)-c) $$

the stationary points (possible solutions) are determined by solving

$$ \nabla f(x^*) +\lambda \nabla g(x^*) = 0 $$

those $x^*$ are the so called stationary points which are potential tangency points between the surfaces $y=f(x)$ and $y = g(x)$.

Suppose now that $f(x) = a_1x_1+a_2x_2+a_3 x_3$ and $g(x) = b_1 x_1+b_2 x_2 + b_3 x_3$. In this case tangency only could occur if $(a_1,a_2,a_3) = k(b_1,b_2,b_3)$.

Concluding, if $f(x)$ and $g(x)$ doesn't have possibilities to be tangent at any point, the lagrangian method is worthless.

Answer 2

Lagrange Multipliers are typically used in constrained optimization, so if you have one constraint, for example, you will enforce 3 constraints tied to the partial derivatives and the actual constraint, yielding you 4 equations for 4 unknowns...