## Agent_A one year ago Lagrange Multipliers Question (See photo) Need to see a complete solution. I want to check it against mine. Thanks!

1. Agent_A

2. anonymous

So you're optimizing $$2x^3+y^4$$ subject to $$x^2+y^2<=1$$. The first step would be to find the critical points of the target function: $\frac{\partial}{\partial x}\left[2x^3+y^4\right]=6x^2\quad\quad\quad\frac{\partial}{\partial y}\left[2x^3+y^4\right]=4y^3$ Setting the partial derivatives equal to $$0$$ indicates that the only critical point occurs at $$(0,0)$$.

3. anonymous

Next you want to find any other extreme values by setting $$\nabla f(x,y)=\lambda \nabla g(x,y)$$, where $$f(x,y)$$ is the function you're optimizing here and $$g(x,y)$$ is the constraint function. $\nabla f(x,y)=\nabla(2x^3+y^4)=\left\langle6x^2,4y^3\right\rangle\\ \lambda\nabla g(x,y)=\lambda\nabla(x^2+y^2)=\lambda\left\langle2x,2y\right\rangle$ Matching up the corresponding components, you have $\begin{cases}6x^2=2\lambda x\\ 4y^3=2\lambda y\end{cases}~~\implies~~\begin{cases}x(3x-\lambda)=0 \\ y(2y^2-\lambda)=0 \end{cases}$ We have several solutions here: $$x=0$$, $$y=0$$, $$x=\dfrac{\lambda}{3}$$, $$y=\sqrt{\dfrac{\lambda}{2}}$$, and $$y=-\sqrt{\dfrac{\lambda}{2}}$$. The point $$(0,0)$$ corresponds to the critical point we already found, so we'll omit that one. Besides, if $$x,y=0$$ then we immediately have $$\lambda=0$$, which doesn't get us anywhere.

4. anonymous

If $$x=0$$, the constraint tells us that $$y^2=1$$, or $$y=\pm1$$. If $$x=\dfrac{\lambda}{3}$$, then $$\dfrac{\lambda^2}{9}+y^2=1$$ gives $$y=\pm\sqrt{1-\dfrac{\lambda^2}{9}}$$. If $$y=0$$, we get $$x=\pm1$$. If $$y=\pm\sqrt{\dfrac{\lambda}{2}}$$, then $$x^2+\dfrac{\lambda}{2}=1$$ gives $$x=\pm\sqrt{1-\dfrac{\lambda}{2}}$$. Plug in the solutions in terms of $$\lambda$$ into the constraint equation. Completing the square and solving gives you two possible values for $$\lambda$$: $\left(\pm\sqrt{1-\dfrac{\lambda}{2}}\right)^2+\left(\pm\sqrt{1-\dfrac{\lambda^2}{9}}\right)^2=1~~\implies~~ \frac{1}{9}\left(\lambda+\frac{9}{4}\right)^2-\frac{25}{16}=0$ which has two roots, $$\lambda=-6$$ and $$\lambda=\dfrac{3}{2}$$. If $$\lambda=-6$$, then $$x=-2$$ and $$y$$ is complex, so we can omit this case. If $$\lambda=\dfrac{3}{2}$$, then $$x=\dfrac{1}{2}$$ and $$y=\pm\sqrt{\dfrac{3}{4}}$$. All this to say we have two points to check: $$\left(\dfrac{1}{2},-\sqrt{\dfrac{3}{4}}\right)$$ and $$\left(\dfrac{1}{2},\sqrt{\dfrac{3}{4}}\right)$$

5. anonymous

Actually, not just those two, we have other cases to consider as well: $\begin{matrix} (0,-1)&&(0,1)&&(1,0)&&(-1,0) \end{matrix}$ So plug all these points into $$f(x,y)=2x^3+y^4$$. Which point gives the highest value? lowest?

6. Agent_A

Thank You, @SithsAndGiggles!