# 10.6 Applications  (Page 2/4)

 Page 2 / 4

## Sample set b

The length of a rectangle is 4 inches more than twice its width. The area is 30 square inches. Find the dimensions (length and width).

Step 1:   Let $x=$ the width. Then, $2x+4=$ the length.

$\begin{array}{lll}\text{Step\hspace{0.17em}}2:\hfill & \hfill & \text{The\hspace{0.17em}area\hspace{0.17em}of\hspace{0.17em}a\hspace{0.17em}rectangle\hspace{0.17em}is\hspace{0.17em}defined\hspace{0.17em}to\hspace{0.17em}be\hspace{0.17em}the\hspace{0.17em}length\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}\hspace{0.17em}rectangle\hspace{0.17em}times\hspace{0.17em}the\hspace{0.17em}width\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}rectangle}\text{.\hspace{0.17em}Thus,}\hfill \\ \hfill & \hfill & x\left(2x+4\right)=30\hfill \end{array}$
$\begin{array}{lllllll}\text{Step\hspace{0.17em}}3:\hfill & \hfill & \hfill x\left(2x+4\right)& =\hfill & 30\hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill 2{x}^{2}+4x& =\hfill & 30\hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill 2{x}^{2}+4x-30& =\hfill & 0\hfill & \hfill & \text{Divide\hspace{0.17em}each\hspace{0.17em}side\hspace{0.17em}by\hspace{0.17em}2}.\hfill \\ \hfill & \hfill & \hfill {x}^{2}+2x-15& =\hfill & 0\hfill & \hfill & \text{Factor}.\hfill \\ \hfill & \hfill & \hfill \left(x+5\right)\left(x-3\right)& =\hfill & 0\hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill x& =\hfill & -5,\text{\hspace{0.17em}}3\hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill x& =\hfill & -5\hfill & \hfill & \text{has\hspace{0.17em}\hspace{0.17em}no\hspace{0.17em}physical\hspace{0.17em}meaning\hspace{0.17em}so\hspace{0.17em}we\hspace{0.17em}disregard\hspace{0.17em}it}\text{.\hspace{0.17em}Check\hspace{0.17em}}x=3.\hfill \\ \hfill & \hfill & \hfill x& =\hfill & 3\hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill 2x+4=2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3+4& =\hfill & 10\hfill & \hfill & \hfill \end{array}$
$\begin{array}{llllll}\text{Step\hspace{0.17em}}4:\hfill & \hfill & \hfill x\left(2x+4\right)& =\hfill & 30\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill 3\left(2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3+4\right)& =\hfill & 30\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill 3\left(6+4\right)& =\hfill & 30\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill 3\left(10\right)& =\hfill & 30\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill 30& =\hfill & 30\hfill & \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct}\text{.}\hfill \end{array}$
$\begin{array}{lll}\text{Step\hspace{0.17em}}5:\hfill & \hfill & \text{Width}=\text{3\hspace{0.17em}inches\hspace{0.17em}and\hspace{0.17em}length}=\text{1}0\text{\hspace{0.17em}inches}.\hfill \end{array}$

## Practice set b

The length of a rectangle is 3 feet more than twice its width. The area is 14 square feet. Find the dimensions.

width = 2 feet, length = 7 feet

The area of a triangle is 24 square meters. The base is 2 meters longer than the height. Find the base and height. The formula for the area of a triangle is $A=\frac{1}{2}b·h.$

height = 6 meters, base = 8 meters

## Sample set c

The product of two consecutive integers is 156. Find them.

$\begin{array}{l}\begin{array}{lllll}\text{Step\hspace{0.17em}1}:\hfill & \hfill & \hfill \text{Let\hspace{0.17em}}x& =\hfill & \text{the\hspace{0.17em}smaller\hspace{0.17em}integer}\text{.}\hfill \\ \hfill & \hfill & \hfill x+1& =\hfill & \text{the\hspace{0.17em}next\hspace{0.17em}integer}\text{.}\hfill \\ \text{Step\hspace{0.17em}2}:\hfill & \hfill & \hfill x\left(x+1\right)& =\hfill & 156\hfill \\ \text{Step\hspace{0.17em}3}:\hfill & \hfill & \hfill x\left(x+1\right)& =\hfill & 156\hfill \\ \hfill & \hfill & \hfill {x}^{2}+x& =\hfill & 156\hfill \\ \hfill & \hfill & \hfill {x}^{2}+x-156& =\hfill & 0\hfill \\ \hfill & \hfill & \hfill \left(x-12\right)\left(x-13\right)& =\hfill & 0\hfill \\ \hfill & \hfill & \hfill x& =\hfill & 12,\text{\hspace{0.17em}}-13\hfill \end{array}\\ \end{array}$
This factorization may be hard to guess. We could also use the quadratic formula.

$\begin{array}{l}{x}^{2}+x-156=0\\ \begin{array}{lllll}a=1,\hfill & \hfill & b=1,\hfill & \hfill & c=-156\hfill \end{array}\\ \begin{array}{lllll}\hfill x& =\hfill & \frac{-1±\sqrt{{1}^{2}-4\left(1\right)\left(-156\right)}}{2\left(1\right)}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{-1±\sqrt{1+624}}{2}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{-1±25}{2}\hfill & \hfill & \frac{-1±25}{2}=\frac{24}{2}=12\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\frac{-1-25}{2}=\frac{-26}{2}=-13\hfill \\ \hfill x& =\hfill & 12,\text{\hspace{0.17em}}-13\hfill & \hfill & \text{Check\hspace{0.17em}12,\hspace{0.17em}13\hspace{0.17em}and\hspace{0.17em}}-\text{13,\hspace{0.17em}12}\text{.}\hfill \\ \hfill x+1& =\hfill & 13,\text{\hspace{0.17em}}-12\hfill & \hfill & \hfill \end{array}\\ \end{array}$ $\begin{array}{l}\begin{array}{rrrrrrrr}\hfill \text{Step}\text{\hspace{0.17em}}4:& \hfill & \hfill \text{If}\text{\hspace{0.17em}}x=12:& \hfill & \hfill 12\left(2+1\right)& \hfill =& \hfill 156& \hfill \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\\ \hfill & \hfill & \hfill & \hfill & \hfill 12\left(13\right)& \hfill =& \hfill 156& \hfill \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\\ \hfill & \hfill & \hfill & \hfill & \hfill 156& \hfill =& \hfill 156& \hfill \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\\ \hfill & \hfill & \hfill \text{If}\text{\hspace{0.17em}}x=-13& \hfill & \hfill -13\left(-13+1\right)& \hfill =& \hfill 156& \hfill \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\\ \hfill & \hfill & \hfill & \hfill & \hfill -13\left(-12\right)& \hfill =& \hfill 156& \hfill \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\\ \hfill & \hfill & \hfill & \hfill & \hfill 156& \hfill =& \hfill 156& \hfill \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct}\text{.}\end{array}\hfill \\ \begin{array}{lllll}\mathrm{Step}\text{\hspace{0.17em}}5:\hfill & \hfill & \text{There}\text{\hspace{0.17em}}\text{are}\text{\hspace{0.17em}}two\text{\hspace{0.17em}}\text{solutions:}\hfill & \hfill & 12,\text{\hspace{0.17em}}13\text{\hspace{0.17em}}\text{and}-13,\text{\hspace{0.17em}}-12.\hfill \end{array}\hfill \end{array}$

## Practice set c

The product of two consecutive integers is 210. Find them.

14 and 15, amd –14 and –15

Four is added to an integer and that sum is tripled. When this result is multiplied by the original integer, the product is −12. Find the integer.

–2

## Sample set d

A box with no top and a square base is to be made by cutting out 2-inch squares from each corner and folding up the sides of a piece of a square cardboard. The volume of the box is to be 8 cubic inches. What size should the piece of cardboard be?

$\begin{array}{lll}\text{Step\hspace{0.17em}1:}\hfill & \hfill & \text{Let\hspace{0.17em}}x=\text{the\hspace{0.17em}length\hspace{0.17em}}\left(\text{and\hspace{0.17em}width}\right)\text{\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}piece\hspace{0.17em}of\hspace{0.17em}cardboard}.\hfill \end{array}$
$\begin{array}{lll}\text{Step\hspace{0.17em}2:}\hfill & \hfill & \text{The\hspace{0.17em}volume\hspace{0.17em}of\hspace{0.17em}a\hspace{0.17em}rectangular\hspace{0.17em}box\hspace{0.17em}is}\hfill \\ \hfill & \hfill & V=\left(\text{length}\right)\text{\hspace{0.17em}}\left(\text{width}\right)\text{\hspace{0.17em}}\left(\text{height}\right)\hfill \\ \hfill & \hfill & 8=\left(x-4\right)\left(x-4\right)2\hfill \end{array}$
$\begin{array}{lllll}\text{Step\hspace{0.17em}}3:\hfill & \hfill & 8=\left(x-4\right)\left(x-4\right)2\hfill & \hfill & \hfill \\ \hfill & \hfill & 8=\left({x}^{2}-8x+16\right)2\hfill & \hfill & \hfill \\ \hfill & \hfill & 8=2{x}^{2}-16x+32\hfill & \hfill & \hfill \\ \hfill & \hfill & 2{x}^{2}-16x+24=0\hfill & \hfill & \text{Divide\hspace{0.17em}each\hspace{0.17em}side\hspace{0.17em}by\hspace{0.17em}2}\text{.}\hfill \\ \hfill & \hfill & {x}^{2}-8x+12=0\hfill & \hfill & \text{Factor}\text{.}\hfill \\ \hfill & \hfill & \left(x-6\right)\left(x-2\right)=0\hfill & \hfill & \hfill \\ \hfill & \hfill & x=6,\text{\hspace{0.17em}}2\hfill & \hfill & \hfill \end{array}$
$x$ cannot equal 2 (the cut would go through the piece of cardboard). Check $x=6.$
$\begin{array}{rrrrrr}\hfill \text{Step\hspace{0.17em}}4:& \hfill & \hfill \left(6-4\right)\left(6-4\right)2& \hfill =& \hfill 8& \hfill \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\\ \hfill & \hfill & \hfill \left(2\right)\left(2\right)2& \hfill =& \hfill 8& \hfill \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\\ \hfill & \hfill & \hfill 8& \hfill =& \hfill 8& \hfill \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct}\text{.}\end{array}$
$\begin{array}{rrr}\hfill \text{Step\hspace{0.17em}}5:& \hfill & \hfill \text{The\hspace{0.17em}piece\hspace{0.17em}of\hspace{0.17em}cardboard\hspace{0.17em}should\hspace{0.17em}be\hspace{0.17em}6\hspace{0.17em}inches\hspace{0.17em}by\hspace{0.17em}6\hspace{0.17em}inches}.\end{array}$

## Practice set d

A box with no top and a square base is to be made by cutting 3-inch squares from each corner and folding up the sides of a piece of cardboard. The volume of the box is to be 48 cubic inches. What size should the piece of cardboard be?

10 in. by 10 in.; 2 by 2 is not physically possible.

## Sample set e

A study of the air quality in a particular city by an environmental group suggests that $t$ years from now the level of carbon monoxide, in parts per million, in the air will be

$A=0.3{t}^{2}+0.1t+4.2$

(a) What is the level, in parts per million, of carbon monoxide in the air now?

Since the equation $A=0.3{t}^{2}+0.1t+4.2$ specifies the level $t$ years from now, we have $t=0.$

$A=0.3{t}^{2}+0.1t+4.2$
$A=4.2$

(b) How many years from now will the level of carbon monoxide be at 8 parts per million?
$\begin{array}{rrrrr}\hfill \text{Step\hspace{0.17em}}1:& \hfill & \hfill t& \hfill =& \hfill \text{the\hspace{0.17em}number\hspace{0.17em}of\hspace{0.17em}years\hspace{0.17em}when\hspace{0.17em}the\hspace{0.17em}level\hspace{0.17em}is\hspace{0.17em}8}.\end{array}$
$\begin{array}{rrrrr}\hfill \text{Step\hspace{0.17em}}2:& \hfill & \hfill 8& \hfill =& \hfill 0.3{t}^{2}+0.1t+4.2\end{array}$
$\begin{array}{lllllll}\text{Step\hspace{0.17em}}3:\hfill & \hfill & 8\hfill & =\hfill & 0.3{t}^{2}+0.1t+4.2\hfill & \hfill & \hfill \\ \hfill & \hfill & 0\hfill & =\hfill & 0.3{t}^{2}+0.1t-3.8\hfill & \begin{array}{l}\text{This\hspace{0.17em}does\hspace{0.17em}not\hspace{0.17em}readily\hspace{0.17em}factor,\hspace{0.17em}so\hspace{0.17em}}\\ \text{we'll\hspace{0.17em}use\hspace{0.17em}the\hspace{0.17em}quadratic\hspace{0.17em}formula}\text{.}\end{array}\hfill & \hfill \\ \hfill & \hfill & a\hfill & =\hfill & 0.3,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}b=0.1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=-3.8\hfill & \hfill & \hfill \\ \hfill & \hfill & t\hfill & =\hfill & \frac{-0.1±\sqrt{{\left(0.1\right)}^{2}-4\left(0.3\right)\left(-3.8\right)}}{2\left(0.3\right)}\hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill & =\hfill & \frac{-0.1±\sqrt{0.01+4.56}}{0.6}=\frac{-0.1±\sqrt{4.57}}{0.6}\hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill & =\hfill & \frac{-0.1±2.14}{0.6}\hfill & \hfill & \hfill \\ \hfill & \hfill & t\hfill & =\hfill & 3.4\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-3.73\hfill & \hfill & \hfill \end{array}$
$\begin{array}{lll}\hfill & \hfill & t=-3.73\text{\hspace{0.17em}}\text{has\hspace{0.17em}no\hspace{0.17em}physical\hspace{0.17em}meaning}.\text{\hspace{0.17em}Check}\text{\hspace{0.17em}}t=3.4\hfill \\ \text{Step\hspace{0.17em}4}:\text{\hspace{0.17em}}\hfill & \hfill & \text{This\hspace{0.17em}value\hspace{0.17em}of\hspace{0.17em}}t\text{\hspace{0.17em}has\hspace{0.17em}been\hspace{0.17em}rounded\hspace{0.17em}to\hspace{0.17em}the\hspace{0.17em}nearest\hspace{0.17em}tenth}.\text{\hspace{0.17em}It\hspace{0.17em}does\hspace{0.17em}check\hspace{0.17em}}\left(\text{pretty\hspace{0.17em}closely}\right).\text{\hspace{0.17em}}\hfill \\ \text{Step\hspace{0.17em}5}:\text{\hspace{0.17em}}\hfill & \hfill & \text{About\hspace{0.17em}3}.\text{4\hspace{0.17em}years\hspace{0.17em}from\hspace{0.17em}now\hspace{0.17em}the\hspace{0.17em}carbon\hspace{0.17em}monoxide\hspace{0.17em}level\hspace{0.17em}will\hspace{0.17em}be\hspace{0.17em}8}.\text{\hspace{0.17em}}\hfill \end{array}$

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