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Let $$h(x,y) = \min\lbrace x,y \rbrace$$ I want to find $$\min ax + by$$ Subject to $$g(x,y) = c - h(x,y)= 0$$ My lagrangian is

$$L(x,y,\lambda) = ax+by + \lambda (c - h(x,y)) $$

I have calculated

$$L_x = a - \lambda h_x(x,y) =0$$ $$L_y = b - \lambda h_y(x,y) =0$$ $$L_\lambda = c - h(x,y) =0$$ So $$\lambda = \frac{a}{h_x} $$ $$\lambda = \frac{b}{h_y} $$

I know the partials for the $\min$ function. But this implies for $x<y$

$$ \frac{a}{1} = \frac{b}{0}$$ and for $x>y$

$$ \frac{a}{0} = \frac{b}{1}$$

I think this implies $\lambda$ is undefined over all intervals.

My Question Have I misused the first order conditions somehow? Why am I getting nonsense answers? Shouldn't the optimum be $x=y$?

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  • $\begingroup$ I am not sure whether you can apply Lagrange multiplier method if $h(x,y)$ does not have continuous partial derivatives. $\endgroup$ – Martin Sleziak Feb 4 '17 at 6:59
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You have for $x>y$, $ a=0, b=\lambda$, and for $x<y$ $ a=\lambda , b=0$. Btw, your third derivative should be $c-h(x,y)=0$ and not $c-h_x =0$. Now write $c=h(x,y)$, for the first case, $c= y$, so the minimum value is $cb$, for the second case the minimum value is $ac$.

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  • $\begingroup$ How do you get $cb$ or $ac$? I don't follow although your conclusion seems to make sense since those values would make sense with the graph I have drawn. $\endgroup$ – Stan Shunpike Apr 26 '15 at 7:53
  • $\begingroup$ For $x>y$ , $h=y$, and subject to the constraint $h=c$ we have $y=c$, so from what we found on the lagrangian we know that a=0, and so the minimum of $ax+by$ is $bc$. For the other case reverse the roles of $a$ and $b$. $\endgroup$ – MathematicalPhysicist Apr 26 '15 at 8:10
  • $\begingroup$ Eccellente. Grazie amico! $\endgroup$ – Stan Shunpike Apr 26 '15 at 8:12

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