### Author Topic: Alternative solution to the optimization in Problem 2 on the practice final  (Read 501 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

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##### Re: Alternative solution to the optimization in Problem 2 on the practice final
« Reply #1 on: April 25, 2022, 07:51:30 AM »
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
It will be, however, more cumbersome. Note that (1) corresponds to two rays and one arc, (2) to two corners.

No, you need not consider quadratic forms after you found all suspicious points. It would serve no purpose.