The starting function I have is f=(1/6)x^3 + 2xy^2 + xy
I need to find the 1st and 2nd partial derivatives, and hence find the location and nture of the four stationary points. How do I go about doing that?
Hi
To find the partial derivative with respect to x of $\displaystyle f(x) = \frac{1}{6}\:x^3 + 2xy^2 + xy$ you must consider y as constant
$\displaystyle \frac{\partial f}{\partial x} = \frac{1}{6}\:3\:x^2 + 2y^2 + y = \frac{1}{2}\:x^2 + 2y^2 + y$
Same method to calculate $\displaystyle \frac{\partial f}{\partial y}$ : consider x as constant
$\displaystyle f=\frac{1}{6}x^3+2xy^2+xy$
Partial differentiation basically involves choosing a variable to differentiate with respect to, and treating all others as constants. So:
$\displaystyle \frac{\partial f}{\partial x}=f_x=\frac{1}{2}x^2+2y^2+y$
$\displaystyle \frac{\partial f}{\partial y}=f_y=4x+x=5x$
I leave you to find
$\displaystyle \frac{\partial ^2f}{\partial x^2}=f_{xx}$
$\displaystyle \frac{\partial ^2f}{\partial x\partial y}=f_{xy}=f_{yx}=\frac{\partial ^2f}{\partial y\partial x}$
$\displaystyle \frac{\partial ^2f}{\partial y^2}=f_{yy}$
Ok I see where you're coming from, as I basically did the same thing before I made this post. I just needed to see if my working as correct. The only problem now for me was to find the stationary points and their nature, i.e. max or min. This is where I am stuck, the first part was pretty easy.
Hello, LooNiE!
To find the stationary points, set the first partials equal to zero and solve.The starting function is: .$\displaystyle f(x,y)\:=\:\tfrac{1}{6}x^3 + 2xy^2 + xy$
I need to find the 1st and 2nd partial derivatives,
and hence find the location and nature of the four stationary points.
How do I go about doing that? . . And no one has told you how?
. . . $\displaystyle \begin{array}{ccccccc}f_x &=&\frac{1}{2}x^2 + 2y^2 + y &=&0 & {\color{red}[1]} \\ \\[-3mm]
f_y &=&4xy + x &=&0 & {\color{red}[2]}\end{array} $
From [2], we have: .$\displaystyle x(4y+1) \:=\:0 \quad\Rightarrow\quad x = 0,\:y = \text{-}\tfrac{1}{4}$
Substitute $\displaystyle x=0$ into [1]: .$\displaystyle 2y^2 + y \:=\:0 \quad\Rightarrow\quad y \:=\:0,\:\text{-}\tfrac{1}{2}$
. . We have two of the points: .$\displaystyle {\color{blue}(0,0),\:\left(0,\text{-}\tfrac{1}{2}\right)} $
Substitute $\displaystyle y = \text{-}\tfrac{1}{4}$ into [1]: .$\displaystyle \tfrac{1}{2}x^2 + \tfrac{1}{8} -\tfrac{1}{4} \:=\:0 \quad\Rightarrow\quad x = \pm\tfrac{1}{2}$
. . And we have two more points: .$\displaystyle {\color{blue}\left(\tfrac{1}{2},\:\text{-}\tfrac{1}{4}\right),\:\left(\text{-}\tfrac{1}{2},\:\text{-}\tfrac{1}{4}\right)} $
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
For the Second Partial Test, form the expression: .$\displaystyle D \;=\;\left(f_{xx}\right)\left(f_{yy}\right) - \left(f_{xy}\right)^2 $
We have: .$\displaystyle f_{xx}\:=\:x,\;\;f_{yy}\:=\:4x,\;\;f_{xy} \:=\:4y+1$
. . Then: .$\displaystyle D \;=\;4x^2 - (4y+1)^2$
At $\displaystyle {\color{blue}(0,0)}\!:\; D\:=\:4(0) - (0+1)^2 \:=\:-1 $ . . . negative: saddle point
At $\displaystyle {\color{blue}\left(0,\text{-}\tfrac{1}{2}\right)}\!:\;D \:=\:4(0) - (\text{-}2+1)^2 \:=\:-1$ . . . negative: saddle point
At $\displaystyle {\color{blue}\left(\tfrac{1}{2},\text{-}\tfrac{1}{4}\right)}\!:\;D\:=\:4(\tfrac{1}{4}) - (-1+1)^2 \:=\:+1$
. . . $\displaystyle f_{xx} \:=\:+\tfrac{1}{2}$ . . . positive, concave up, $\displaystyle \cup$ . . . minimum
At $\displaystyle {\color{blue}\left(\text{-}\tfrac{1}{2},\text{-}\tfrac{1}{4}\right)}\!:\;D \:=\:4(\tfrac{1}{4}) - (-1+1)^2 \:=\:+1$
. . . $\displaystyle f_{xx} \:=\:\text{-}\tfrac{1}{2}$ . . . negative, concave down, $\displaystyle \cap$ . . . maximum