243.14.7.4 Find all local extreme values

[karush]

$\tiny{243.14.7.4}$
Find all local extreme values of the given function and
identify each as a local maximum,local minimum,or saddlepoint
\begin{align*} \displaystyle
f_4(x,y)&=x^2+14x+y^2-12y+5\\
\end{align*}

saw this in the section

$\text{if we substitute the values} \\$
$\text{$f_xCa, b) = 0$ and $f_y(a, b) = 0$}\\$
$\text{ into the equation both zero.}\\$
$\text{ $f_x(a, b)(x - a)+f_y(a, b)(y - b) - (z - f(a, b)) = 0$ } $

ok i haven't done this before so kinda lost
read the section on but didn't get the derivatives
answer
$\color{red}{f_4(-7,6)}=\color{red}{-80}\\$
$\color{red}{\textit{local minimum}}$

 

[HallsofIvy]

$f(x,y)= x^2+ 14x+ y^2- 12y+ 5$. The partial derivatives are $f_x= 2x+ 14$ and $f_y= 2y- 12$. Setting those equal to 0, $2x+ 14= 0$ gives $x= -7$ and $2y- 12= 0$ gives $y= 6$. The only possible extremum is, as you say, at (-7, 6). I have no idea what the subscript "4" indicates. Was that a typo?

There are several different ways to determine whether that extremum is a "maximum", "minimum", or "saddle point".

One is to look at the second derivatives. $f_{xx}= 2$, $f_{yy}= 2$, and $f_{xy}= 0$. Those are all positive meaning that the extremum is convex upward so a minimum.

And one way to do this without using derivatives at all is to "complete the square" in both x and y: $f(x, y)= x^2+ 14x+ 49- 49+ y^2- 12y+ 36- 36+ 5= (x- 7)^2+ (y- 6)^2- 49- 36+ 5= (x+ 7)^2+ (y- 6)^2- 80$. That last shows that the graph is a paraboloid with vertex at (-7, 6, -80) opening upward.

 

[karush]

subscipt 4 is the problem number
but probably shouldn't use it with these.

ok I have more of these to do
trying to finish up this homework set before end of year
trying to get into calc IV