10.6 Applications  (Page 2/4)

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.    

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

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

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

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

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?

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

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

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

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

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

   $A=0.3t^2+0.1t+4.2$
   $A=4.2$

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