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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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 \((0,0)\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Next 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If \(x=0\), 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 \(\left(\dfrac{1}{2},\sqrt{\dfrac{3}{4}}\right)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Actually, 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?

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.0Thank You, @SithsAndGiggles!
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