MHB -x4.6 Find extrema f(x, y, z) = x + yz

[karush]

$\text{Find the extrema of $f(x, y, z) = x + yz$ on the line defined by}$
$$\text{$x = 8(2 + t), y = t - 8,$ and $z = t+ 2$.}$$
$\text{Classify each extremum as a minimum or maximum.}$
\begin{align*} \displaystyle
&
\text{Book answer}=\color{red}{\text{$(8, -9, 1)$, minimum}}
\end{align*}
$\textit{ok not sure how you take derivative of $f(x,y,z)$ to set it to zero}$
$\tiny{x4.6}$

 

[S.G. Janssens]

Hint: If you substitute the expressions for $x, y$ and $z$ in terms of $t$ into the expression for $f(x,y,z)$, then you get a single-variable function of $t$.

 

[karush]

thusly..:D

\begin{align*} \displaystyle
f_6(x,y,z)&=x+yz\\
f_6(8(2+t),t-8,t+2)&=8(2+t)+(t-8)(t+2)\\
&=t^2+2t\\
f_6^\prime(-1)&=0\\
&=8(2+(-1))+((-1)-8)((-1)+2)\\
&=8+(-9)(1)=x+yz\\
\therefore&=\color{red}
{\text{$(8,-9,1)$,min}}
\end{align*}

 

[HallsofIvy]

OR you can use the "chain rule"- if f is a function of x, y, and z and x, y, and z are functions of t then $$\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}+ \frac{\partial f}{\partial z}\frac{dz}{dt}$$.

Since [math]f(x, y)= x+ yz[/math], [math]\frac{\partial f}{\partial x}= 1[/math], [math]\frac{\partial f}{\partial y}= z[/math], and [math]\frac{\partial f}{\partial z}= y[/math].

Since [math]x= 8(2+ t)= 16+ 8t[/math], [math]\frac{dx}{dt}= 8[/math]. Since [math]y= t- 8[/math], [math]\frac{dy}{dt}= 1[/math]. Since [math]z= t+ 2[/math], [math]\frac{dz}{dt}= 1[/math].

Putting those all together, [math]\frac{df}{dt}= (1)(8)+ (z)(1)+ (y)(1)= 8+ z+ y= 8+ t+ 2+ t- 8= 2t+ 2[/math]. That will be 0 when t= -1 so when x= 16- 8= 8, y= -1- 8= -9, and z= -1+ 2= 1. f(8, -9, 1)= 8- 9= -1.

 

[karush]

https://dl.orangedox.com/GXEVNm73NxaGC9F7Cy