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

• MHB
• karush
In summary, the extrema of $f(x, y, z) = x + yz$ on the line defined by $x = 8(2 + t), y = t - 8,$ and $z = t+ 2$ are found by using the chain rule and setting the derivative of $f$ with respect to $t$ equal to 0. This results in the point $(8, -9, 1)$ being a minimum for the function.
karush
Gold Member
MHB
$\text{Find the extrema of$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)$to set it to zero}$
$\tiny{x4.6}$

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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 $t$.

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*}

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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 $$\displaystyle \frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}+ \frac{\partial f}{\partial z}\frac{dz}{dt}$$.

Since $$\displaystyle f(x, y)= x+ yz$$, $$\displaystyle \frac{\partial f}{\partial x}= 1$$, $$\displaystyle \frac{\partial f}{\partial y}= z$$, and $$\displaystyle \frac{\partial f}{\partial z}= y$$.

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

Putting those all together, $$\displaystyle \frac{df}{dt}= (1)(8)+ (z)(1)+ (y)(1)= 8+ z+ y= 8+ t+ 2+ t- 8= 2t+ 2$$. 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.

https://dl.orangedox.com/GXEVNm73NxaGC9F7Cy

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## 1. What is the purpose of finding the extrema for a function?

The extrema of a function represent the maximum and minimum values of the function. These points can provide valuable information about the behavior and properties of the function, such as its critical points and overall shape.

## 2. How do you find the extrema of a multi-variable function like f(x, y, z) = x + yz?

To find the extrema of a multi-variable function, we use partial derivatives. We take the partial derivative of the function with respect to each variable and set them equal to zero. We then solve the resulting system of equations to find the critical points, which can be potential extrema.

## 3. What is the difference between a local and global extremum?

A local extremum is a point where the function reaches its maximum or minimum value within a specific region, but it may not be the overall maximum or minimum for the entire function. A global extremum, on the other hand, is the absolute maximum or minimum value of the function for its entire domain.

## 4. How can we determine if a critical point is a local maximum or minimum?

To determine if a critical point is a local maximum or minimum, we use the second derivative test. We take the second partial derivatives of the function and evaluate them at the critical point. If the resulting value is positive, then the critical point is a local minimum. If it is negative, then the critical point is a local maximum.

## 5. Can a function have multiple extrema?

Yes, a function can have multiple extrema. In fact, a function can have multiple local and global extrema within its domain. It is important to consider both the first and second derivatives of a function to accurately identify all of its extrema points.

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