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In optimization problem using Lagrange multiplier..

$f(x, y, z) = x + 2z$, subject to $g_1(x, y, z) = x + y + z = 1$, $g_2(x; y; z) = 2x + z = 2$

Find $x, y, x, \lambda_1, \lambda_2$

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I couldn't find the right solution using with equation $ℒ = f(x)-\lambda_1 g_1(x, y, z)-λ_2g_2(x, y, z)=0$ Should I have doubts about the problem? Thank you for reading.

Stackman
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woogawooga
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  • Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. – José Carlos Santos Oct 07 '18 at 06:51
  • You might try graphing the function and constraints to understand why there are no solutions. – Narlin Oct 07 '18 at 10:59

2 Answers2

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For the first case. let $$f(x,y,z,\lambda_1,\lambda_2)=x+2z+\lambda_1(x+y+z-1)+\lambda_2(2x+z-2)$$ then we get

$$\frac{\partial f}{\partial x}=1+\lambda_1+2\lambda_2$$ $$\frac{\partial f}{\partial y}=\lambda_1$$ $$\frac{\partial f}{\partial z}=2+\lambda_1+\lambda_2$$ $$\frac{\partial f}{\partial \lambda_1}=x+y+z-1$$ $$\frac{\partial f}{\partial \lambda_2}=2x+z-2$$ The system has no solutions, since we get $$\lambda_2=-\frac{1}{2}$$ and $$\lambda_2=-2$$

we must solve $$\frac{\partial f}{\partial x}=0$$ $$\frac{\partial f}{\partial y}=0$$ $$\frac{\partial f}{\partial z}=0$$ $$\frac{\partial f}{\partial \lambda_1}=0$$ $$\frac{\partial f}{\partial \lambda_2}=0$$

Dr. Sonnhard Graubner
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The two constraints define a feasible set which is a line

$$ g_1(x,y,z)\cap g_2(x,y,z) = L $$

$$ L\to (x,y,z)^{\dagger} = (1,0,0)^{\dagger}+\mu(1,1,-2)^{\dagger} $$

so along $L$ the objective function which is linear, is unbounded.

Cesareo
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