For my vector calculus class, I need to solve this problem:
Let $$f(x,y)=x^3+8xy-8y^3-7x+14y$$
Given that this function has critical points on the lines $3x=7$ and $3x=-3$, find and classify all critical points.
MY SOLUTION
Since we have some conditions on our critical points, I decided to use Lagrange multipliers method.
Thus, we have $\nabla f=\lambda \nabla g$ and, depending on which condition we choose, $g_{1}=3x-7=0$ or $g_{2}=3x+3=0$
I, then, need to find the partial derivatives of $f$, $g_{1}$, $g_{2}$: $$\frac{\partial f}{\partial x}=3x^2+8y-7$$ $$\frac{\partial f}{\partial y}=8x-24y^2+14$$ $$\frac{\partial g_{1}}{\partial x}=3$$ $$\frac{\partial g_{1}}{\partial y}=0$$ $$\frac{\partial g_{2}}{\partial x}=3$$ $$\frac{\partial g_{2}}{\partial y}=0$$
For our first condition ($g_{1}$), we have this system of equations: $$\begin{cases}3x^2+8y-7=3\lambda\\8x-24y^2+14=0\\3x-7=0\end{cases}$$
$\Rightarrow x=\frac{7}{3}$
$\Rightarrow y=\pm \sqrt{\frac{8x+14}{24}}$
$\Rightarrow y=\pm \sqrt{\frac{8\cdot \frac{7}{3}+14}{24}}=\pm \frac{7}{6}$
Using the first condition gave me 2 critical points: $(\frac{7}{3},\frac{7}{6})$ and $(\frac{7}{3},-\frac{7}{6})$
So, now, let's use our second condition ($g_{2}$) to find some other solutions:
$$\begin{cases}3x^2+8y-7=3\lambda\\8x-24y^2+14=0\\3x+3=0\end{cases}$$
$\Rightarrow x=-1$
$\Rightarrow y=\pm \sqrt{\frac{8x+14}{24}}$
$\Rightarrow y=\pm \sqrt{\frac{8\cdot -1+14}{24}}=\pm \frac{1}{2}$
Using the second condition gave me 2 other critical points: $(-1,\frac{1}{2})$ and $(-1,-\frac{1}{2})$
So now that I know what are the critical points, I want to classify them (local maximum, local minimum, saddle point). And, from what I remember from Calculus III, the rules to know the category of each point were:
If $D>0$ and $f_{xx}>0$, then it's a minimum
If $D>0$ and $f_{xx}<0$, then it's a maximum
If $D<0$, then it's a saddle point
If $D=0$, then we cannot determine
where $D=f_{xx}f_{yy}-f_{xy}^2$
But, in this course, we have to compute the Hessian matrix and do something with the determinant and eigenvalues. I can see that the determinant of the Hessian matrix is the "D" I was using in Calc III but what about a $3\times 3$ matrix. If someone could explain it to me, it would be great.
Anyway, I calculated the Hessian and its determinant D: $$\begin{bmatrix} \frac{\partial^2 f}{\partial x^2}&\frac{\partial^2 f}{\partial x\partial y}\\\frac{\partial^2 f}{\partial y\partial x}&\frac{\partial^2 f}{\partial y^2}\end{bmatrix}=\begin{bmatrix} 6x&8\\8&-48y\end{bmatrix}$$ $$D=f_{xx}f_{yy}-f_{xy}^2=6x\cdot -48y-8^2=-288xy-64$$
I, then, calculated D for each of my points and I concluded this, using what I learned in Calc III: $$D_{(\frac{7}{3},\frac{7}{6})}=-288\cdot \frac{7}{3}\cdot \frac{7}{6}-64=-848\Rightarrow (\frac{7}{3},\frac{7}{6})\mbox{ is a saddle point}$$ $$D_{(\frac{7}{3},-\frac{7}{6})}=-288\cdot \frac{7}{3}\cdot -\frac{7}{6}-64=720\Rightarrow (\frac{7}{3},-\frac{7}{6})\mbox{ is a local minimum}$$ $$D_{(-1,\frac{1}{2})}=-288\cdot -1\cdot \frac{1}{2}-64=80\Rightarrow (-1,\frac{1}{2})\mbox{ is a local maximum}$$ $$D_{(-1,-\frac{1}{2})}=-288\cdot -1\cdot -\frac{1}{2}-64=-208\Rightarrow (-1,-\frac{1}{2})\mbox{ is a saddle point}$$
But when I entered my answers into Webwork to have them validated, it told me that my maximum and minimum were correct but my two saddle points were wrong. So I asked my teacher to check for errors in Webwork and he said this: "Not all answers are correct, please check the condition $f_{x}=0$" but I didn't understand what that meant.
Can you tell me where my mistake is and how my teacher's hint could help me?
Thanks in advance for your help.
IsaacM
My answer to @Andrei:
Do we really have to use $f_{x}=0$ ? Because I feel like it's not always true when we work with a constraint. For example, we want to optimize $f(x,y)=xy+1$ and $g(x,y)=x^2+y^2-1$. We can see on the graph of the function that, when $g(x,y)$ is max $f_{x}$ is not zero. How is it the case ?
Best Answer
What the teacher asked you to do is to check if you really have a critical point. You get $x$ from the "constraint", $y$ from the condition that $\frac{\partial f}{\partial y}=0$, but you did not check that $\frac{\partial f}{\partial x}=0$. Let's look at the $(7/3,7/6)$ point. $$\frac{\partial f}{\partial x}=3x^2+8y-7=3\frac{49}{9}+8\frac76-7=\frac{49+28-21}3\ne 0$$ You need to check for all your points.