# Finding critical points and maxima and minima for multivariable functions

This problem has given me a lot of difficulties. It shouldn't be hard, as I know how to do part A by just taking the partial derivatives with respect to x and y and solving those systems of equations to find the critical points. However, I have completely no idea how to do part b or part c for that matter.

• You should read about a theorem that talks about the hessian matrix(its the matrix that consists of the second partial derivatives of f) – al.al. Mar 25 '18 at 18:59
• In the future, please take the time to enter the content of your question as text instead of posting a picture of it. It’s only fair that you should take some of your own time to formulate the question if you expect others to spend their time to help you. Images are neither searchable nor accessible to people who use screen readers. Use MathJax to format your mathematical expressions; you can find a quick reference here. – amd Mar 26 '18 at 2:37

## 2 Answers

solve the equation System $$3x^2-6x+y+1=0$$ and $$x-2y+1=0$$

For part b. it really just wants you to check all of the critical points and check it with the boundaries to find the absolute max/min, just like in a function of one variable. To check the boundaries, you need to find the four functions that represent each edge of $S$. To do this, we find the intersections: \begin{aligned}f(x,y)&=x^3-3x^2+xy-y^2+x+y\\f(0,y)&=-y^2+y && \text{Bottom edge of }S\\f(2,y)&=-y^2+3y-2 && \text{Top edge of }S\\f(x,0)&=x^3-3x^2+x && \text{Left edge of }S\\f(x,2)&=x^3-3x^2+3x-2 && \text{Right edge of }S\end{aligned} Now, we have 4 single-variable functions to represent the boundaries of $S$. If we find the maxima and minima of these, we can test it with our critical points. We can't forget to find the boundaries of these functions too, which are really just the 4 corners of $S$. Let's find them all:

### Bottom Edge:

\begin{aligned}\frac{d}{dy}\left(-y^2+y\right)=-2y+1=0\to (y,z)&=\left(\frac12,\frac14\right)\\y=0\to(y,z)&=(0,0)\\y=2\to(y,z)&=(2,-2)\end{aligned}

### Top Edge:

\begin{aligned}\frac{d}{dy}\left(-y^2+3y-2\right)=-2y+3=0\to (y,z)&=\left(\frac32,\frac14\right)\\y=0\to(y,z)&=(0,-1)\\y=2\to(y,z)&=(2,0)\end{aligned}

### Left Edge:

\begin{aligned}\frac{d}{dx}\left(x^3-3x^2+x\right)=3x^2-6x+1=0\to(x,z)&=\left(\frac{3+\sqrt{6}}{3},\frac{-9-4\sqrt{6}}{9}\right)\\3x^2-6x+1=0\to(x,z)&=\left(\frac{3-\sqrt{6}}{3},\frac{-9+4\sqrt{6}}{9}\right)\\x=0\to(x,z)&=(0,0)\\x=2\to(x,z)&=(2,-2)\end{aligned}

### Right Edge:

\begin{aligned}\frac{d}{dx}\left(x^3-3x^2+3x-2\right)=3x^2-6x+3=0\to(x,z)&=\left(1,-1\right)\\x=0\to(x,z)&=(0,-2)\\x=2\to(x,z)&=(2,0)\end{aligned} At this point all we have to do is compare the maxima/minima with the values we found on these boundaries. Part c. requires you to find the determinant of the Hessian matrix, below: D = \text{det}\left(\left[\begin{aligned}\frac{\partial^2 f}{\partial x^2} && \frac{\partial^2 f}{\partial x y}\\\frac{\partial^2 f}{\partial y x} && \frac{\partial^2 f}{\partial y^2}\end{aligned}\right]\right) The theorem states if $D>0$ and $f_{xx}>0$, then its a local minimum; if $D>0$ and $f_{xx}<0$, then its a local maximum; if $D<0$, then its a saddle point; and if $D=0$, then there is no conclusion.