# Find $x_1,x_2$ such that $\min_{x_1,x_2} x_1^2+2x_1x_2$ where $x_1,x_2$ are subject to constraint $x_1^2x_2 \ge 10$.

I am attempting to solve a constrained optimization problem using Lagrange multipliers but am getting lost on how to resolve the equations the gradients output.

The problem is the following:

Find $$x_1,x_2$$ such that $$\min_{x_1,x_2} x_1^2+2x_1x_2$$ where $$x_1,x_2$$ are subject to constraint $$x_1^2x_2 \ge 10$$.

I have changed the constraint into the equality $$x_1^2x_2-10-s^2=0$$ and attained the gradients which result in 4 equations and 4 unknowns:

\begin{align} x^2_1x_2-10-s^2 &= 0 \\ 2x_1+2x_2 &= \lambda (2x_1x_2) \\ 2x_1 &= \lambda x_1^2 \\ 0 &= \lambda(-2s) \end{align}

But I am unsure of how to proceed from here. Additionally, I am struggling to find the dual problem.

• If you assume $x_1 \ge 0$ then that should be stated in the question, otherwise there is no minimum.
– dxiv
Commented Apr 3, 2022 at 22:58

The fourth equation implies that $$\lambda=0$$ or $$s=0$$. Suppose $$\lambda=0$$. Then the third equation implies that $$x_1=0$$, which contradicts the original constraint. So $$s=0$$. Because $$x_1 \not= 0$$, the third equation yields $$x_1 = 2 / \lambda$$, which reduces the second equation to $$4/\lambda + 2x_2 = 4x_2,$$ which yields $$x_2 = 2/\lambda = x_1$$. Now the first equation becomes $$x_1^3 = 10$$, which means that $$x_1=x_2=\sqrt[3]{10}$$.
Clearly $$x_1 \neq 0$$ since $$x_1^2x_2 \ge 10$$. Thus $$2 = \lambda x_1\implies 2x_1+2x_2=4x_2\implies x_1=x_2$$. Since $$\lambda \neq 0 \implies s = 0 \implies x_1^3 = 10\implies x_1 = \sqrt[3]{10} = x_2$$. Thus $$\text{min}(x_1^2+2x_1x_2)= 3x_1^2=3\sqrt[3]{100}$$.
$$x_2 \ge \dfrac{10}{x_1^2}\implies x_1^2+2x_1x_2 \ge x_1^2+\dfrac{20}{x_1}=f(x_1)$$. By using first derivative, you can get the same answer as the Lagrange Multiplier method.