I am new to Lagrange multipliers. Could some one show me how to minimize the following function: \begin{align} f(x,y)=ax+by-\sqrt{cxy} \end{align} subject to: \begin{align} 0 &\le x\\ 0 &\le y \\ ax &\le k y \end{align}

  • $\begingroup$ Since the variables are positive, it'll be convenient to define $x=z^2$ and $y=w^2$ to avoid the square roots. Also, those last two constraints are peculiar: what is $k$? $\endgroup$ – Semiclassical Aug 19 '14 at 16:19
  • $\begingroup$ Can you at least setup Lagrange's equations? $\endgroup$ – Alex R. Aug 19 '14 at 16:20
  • $\begingroup$ @Semiclassical k is some constant. Say greater than 1. $\endgroup$ – Boby Aug 19 '14 at 16:21
  • $\begingroup$ @AlexR. No.I can't. Sorry. I newer seen Lagrange optimization with inequalities $\endgroup$ – Boby Aug 19 '14 at 16:22
  • $\begingroup$ The reason it's odd that you've really got two cases, since either $a x \leq b y \leq ky$ or $ax \leq k y \leq by$. (And since the former case is equivalent to the latter for $k=b$, you might as well on $ax\leq k y$.) $\endgroup$ – Semiclassical Aug 19 '14 at 16:25


When a constrain in an inequality ($\leq,\geq$) you should do the following:

  1. Find the extrema of the function, assuming there's no contrain, and retain all points that meet the constrain
  2. Then, use Lagrange multipliers to find all extrema in the contrain.

If we proceed that way in this problem, we'll have the following:

1) Let's find the critical points of $f(x,y)$ that meet $ax<ky$: $$\begin{array}{rcl} \nabla f(x,y)&=&\vec{0}\\ \Rightarrow a-\frac{\sqrt{cy}}{2\sqrt{x}}&=&0\\ b-\frac{\sqrt{cx}}{2\sqrt{y}}&=&0\\ &&\\ \Rightarrow 2a\sqrt{x}-\sqrt{cy}&=&0\\ -\sqrt{cx}+2b\sqrt{y}&=&0\\ \end{array}$$

Using substitution $u=\sqrt{x},\,v=\sqrt{y}$: $$\begin{array}{rcl} \Rightarrow 2au-\sqrt{c}v&=&0\\ -\sqrt{c}u+2bv&=&0\\ \end{array}$$

The linear equation system has determinant $4ab-c$. You'll need conditions on $a,b,c$ to verify if the system has only one solution $4ab\neq c$) or infinite ($4ab=c$).

a) If the system has only one solution, it is obviously $x=y=0$.

b) If the system has infinite solutions, the solutions are given by the base $(\frac{2a}{\sqrt{c}},1)_{u,v}=(\frac{4a^2}{c},1)_{x,y}$ intersecting the region $ax<ky$.

2) Now we use Lagrange multipliers with $ax-ky=0$: $$\begin{array}{rcl} L(x,y,\lambda)&=&f(x,y)-\lambda g(x,y)\\ &=&ax+by-\sqrt{cxy}-\lambda(ax-ky)\\ &=&ax+by-\sqrt{cxy}-\lambda ax-\lambda ky \end{array}$$

Then $$\begin{array}{rclcc} \nabla L(x,y,\lambda)&=&0&&\\ \Rightarrow \nabla f(x,y)&=&\lambda\nabla g(x,y)&&\\ \Rightarrow a-\frac{\sqrt{cy}}{2\sqrt{x}}&=&\lambda a&&(1)\\ b-\frac{\sqrt{cx}}{2\sqrt{y}}&=&-\lambda k&&(2)\\ ax-ky&=&0&&(3)\\ \end{array}$$

Using $(3)$ in $(2)$, $$\begin{array}{rclcc} 2a(1-\lambda)\sqrt{x}&=&\sqrt{c}\sqrt{y}&&(4)\\ 2\biggl(b+\lambda\frac{ax}{y}\biggr)\sqrt{y}&=&\sqrt{c}\sqrt{x}&&(5)\\ \end{array}$$

I think you can solve the equation system from here.

I hope you find this useful. I also think that you would need to impose some assumptions on $a,b,c,k$ to get "closed" solutions. Good luck!


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