Question

Lagrange Multipliers Question

(See photo)

Need to see a complete solution. I want to check it against mine. Thanks!

Answer 1

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 \((0,0)\).

Answer 2

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.

Answer 3

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 \(\left(\dfrac{1}{2},\sqrt{\dfrac{3}{4}}\right)\)

Answer 4

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?

Answer 5

Thank You, @SithsAndGiggles!