### Author Topic: Alternative solution to the optimization in Problem 2 on the practice final  (Read 162 times)

#### Weihan Luo

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##### Alternative solution to the optimization in Problem 2 on the practice final
« on: April 23, 2022, 02:19:38 PM »
Could I have solved the maximization/minimization using Lagrange multipliers? In particular, define $g_1(x,y) = y-x$, $g_2(x,y) = y+x$, and $g_3(x,y) = -(x^2+y^2)+1$. Then, a solution $(x^*,y^*)$ necessarily satisfies $$\nabla{u} + \lambda_1\nabla{g_1} + \lambda_2\nabla{g_2} + \lambda_3\nabla{g_3} = 0$$ and $$\lambda_1{g_1} = 0, \lambda_2{g_2}=0, \lambda_3{g_3}=0$$

for some $\lambda_{i} \geq 0$.

Then, after finding the points $(x^*, y^*)$, I need to verify that $$\nabla^2{u} + \lambda_1\nabla^2{g_1} + \lambda_2\nabla^2{g_2} + \lambda_3\nabla^2{g_3}$$ is positive definite on the tangent space $T_{x^*,y^*}D$.

Would this approach also work?

#### Victor Ivrii

Yes, it can be solved using Lagrange multiplies. However note, if restrictions are $g_1\le 0$, $g_2\le 0$, $g_3\le 0$ you need to consider
• $g_1=0$ (and $g_2\le 0, g_3\le 0)$); there will be only one Lagrange multiplier at $g_1$. Two other cases in the similar way
• $g_1=g_2=0$ (and $g_3\le 0)$); there will be  two Lagrange multipliers. Two other cases in the similar way