# Suppose that f(x, y, z) = (x - 1)^2 + (y - 1)^2 + (z - 1)^2 with 0 \leq x, y, z and x + y + z...

## Question:

Suppose that {eq}f(x, y, z) = (x - 1)^2 + (y - 1)^2 + (z - 1)^2 {/eq} with {eq}0 \leq x, y, z {/eq} and {eq}x + y + z \leq 10 {/eq}.

a. If the critical point of f(x, y, z) is at (a, b, c), find a, b, c.

b. Find the absolute minimum and absolute maximum of f(x, y, z).

## Critical Points and Absolute Extrema:

The points at which the function start increasing or decreasing are called Critical points, the highest value of the function is known as absolute maxima and the lowest value of the function is known as absolute minima

## Answer and Explanation: 1

**Given:**

{eq}f\left( {x,y,z} \right) = {\left( {x - 1} \right)^2} + {\left( {y - 1} \right)^2} + {\left( {z - 1} \right)^2} {/eq}

Substitute {eq}z = 10 -x - y {/eq}

{eq}\begin{align*} f\left( {x,y} \right) &= {\left( {x - 1} \right)^2} + {\left( {y - 1} \right)^2} + {\left( {10 - x - y - 1} \right)^2}\\ &= {\left( {x - 1} \right)^2} + \left( {y - 1} \right) + \left( {9 - x - y} \right) \end{align*} {/eq}

**(a)**

Partially differentiate with respect to {eq}x {/eq} and {eq}y {/eq} ,

{eq}\begin{align*} {f_x} &= 2\left( {x - 1} \right) - 2\left( {9 - x - y} \right)\\ {f_y}& = 2\left( {y - 1} \right) - 2\left( {9 - x - y} \right) \end{align*} {/eq}

Equate to zero,

{eq}\begin{align*} 2\left( {x - 1} \right) - 2\left( {9 - x - y} \right)& = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\ 2\left( {y - 1} \right) - 2\left( {9 - x - y} \right)& = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right) \end{align*} {/eq}

By solving the equation 1 and 2,

{eq}\begin{align*} x &= \dfrac{{10}}{3}\\ y &= \dfrac{{10}}{3} \end{align*} {/eq}

And the value of {eq}z {/eq} is,

{eq}\begin{align*} z &= 10 - x - y\\ &= 10 - \dfrac{{10}}{3} - \dfrac{{10}}{3}\\ &= \dfrac{{10}}{3} \end{align*} {/eq}

Hence the critical points are {eq}\left( {a,b,c} \right) = \left( {\dfrac{{10}}{3},\dfrac{{10}}{3},\dfrac{{10}}{3}} \right) {/eq} .

**(b)**

All the endpoints are lies between {eq}\left( {0,0,0} \right) to\left( {\dfrac{{10}}{3},\dfrac{{10}}{3},\dfrac{{10}}{3}} \right) {/eq}.

And the given function is symmetric so {eq}\left( {0,0,0} \right) {/eq} gives the absolute minimum and {eq}\left( {\dfrac{{10}}{3},\dfrac{{10}}{3},\dfrac{{10}}{3}} \right) {/eq} gives absolute maximum,

The value of absolute minimum is,

{eq}\begin{align*} f\left( {0,0,0} \right)& = {\left( {0 - 1} \right)^2} + {\left( {0 - 1} \right)^2} + {\left( {0 - 1} \right)^2}\\ &= 1 + 1 + 1\\ &= 3 \end{align*} {/eq}

And the value of absolute maximum is,

{eq}\begin{align*} f\left( {\dfrac{{10}}{3},\dfrac{{10}}{3},\dfrac{{10}}{3}} \right) &= {\left( {\dfrac{{10}}{3} - 1} \right)^2} + {\left( {\dfrac{{10}}{3} - 1} \right)^2} + {\left( {\dfrac{{10}}{3} - 1} \right)^2}\\ &= {\left( {\dfrac{7}{3}} \right)^2} + {\left( {\dfrac{7}{3}} \right)^2} + {\left( {\dfrac{7}{3}} \right)^2}\\ &= \dfrac{{49}}{3} \end{align*} {/eq}

#### Learn more about this topic:

from

Chapter 5 / Lesson 2One of the most important practical uses of higher mathematics is finding minima and maxima. This lesson will describe different ways to determine the maxima and minima of a function and give some real world examples.