Exercise 1.1.65 (65-70)

Question

Find a formula for the described function and state its domain.
    \begin{enumerate}
\item[65.] A rectangle has perimeter 20 m. Express the area of the rectangle as a function of the length of one of its sides.
\item[66.] A rectangle has area 16 m2. Express the perimeter of the rectangle as a function of the length of one of its sides.
\item[67.] Express the area of an equilateral triangle as a function of the length of a side.
\item[68.] A closed rectangular box with volume 8 ft3 has length twice the width. Express the height of the box as a function of the width.
\item[69.] An open rectangular box with volume 2 m3 has a square base. Express the surface area of the box as a function of the length
of a side of the base.
\item[70.] A right circular cylinder has volume of 25 in3. Express the radius of the cylinder as a function of the height.
\end{enumerate}

Khagan Gulusoy · Accepted Answer

\begin{enumerate}
    \item [65.] Assume that $l, w\in \mathbb{R}$ are the length and width of the rectangle. Since the perimeter is equal to 20, we have:
    $$2l+2w=20\quad (\div 2)$$
    $$l+w=10$$
    \begin{equation}\label{one}w=10-l\end{equation}
    Therefore, the function $S$ of the area looks like
    $$S(l)=l(10-l)=10l-l^2.$$
    Because both length and width must be positive, from the equation ef{one} we see that $l$ must be positive and smaller than 10, i.e. 
    $$\forall l\in (0,10):\quad S(l)=10l-l^2.$$

\item[66.]  Assume that $l, w\in \mathbb{R}$ are the length and the width of the rectangle. Since the area is equal to 16, we have:
    $$lw=16$$
    \begin{equation}\label{eq2}
        w=\frac{16}{l}.
    \end{equation}
    Therefore, the function $P$ of perimeter looks like:
    $$P(l)=\frac{32}{l}+2l.$$
    Since the length and the width of the rectangle must be positive, from ef{eq2} we have:
    $$\forall l\in (0,+\infty):\quad P(l)=\frac{32}{l}+2l.$$

\item[67.] Let $S$ be the area function of the equilateral triangle with the side length $l$.
    \begin{figure}[H]
        \centering
    \begin{tikzpicture}[scale=0.8]
        % draw the background
        \draw [line width=1.5pt, fill=gray!2] (0,0) -- (60:4) -- (4,0) -- cycle;
        
ewcommand{	hiscolor}{gray!10}
        % points
        \coordinate[label=left:$A$]  (A) at (0,0);
        \coordinate[label=right:$B$] (B) at (4,0);
        \coordinate[label=above:$C$] (C) at (2,3.464);
        % sides
        \coordinate[label=below:$l$](c) at ($ (A)!.5!(B) $);
        \coordinate[label=left:$l$] (b) at ($ (A)!.5!(C) $);
        \coordinate[label=right:$l$](a) at ($ (B)!.5!(C) $);
        % height
        \draw[dashed] (C) -- (2,0);
        \coordinate[label=left:$h$,color=	hiscolor] (h) at (2,3.464/2);
        % angle alpha (left bottom)
        %\draw[fill=	hiscolor] (0,0) -- (0:0.75cm) arc (0:60:.75cm); % mark angles via arc segments
        \draw (0.7cm,0.3cm) node {$60^{\circ}$};
        % angle beta (right bottom)
        \begin{scope}[shift={(4cm,0cm)}]
            %\draw[fill=	hiscolor] (0,0) -- (-180:0.75cm) arc (180:120:0.75cm); % mark angles via arc segments
            \draw (150:0.7cm) node {$60^{\circ}$};
        \end{scope}
        % angle gamma (top)
        \begin{scope}[shift={(60:4)}]
            %\draw[fill=	hiscolor] (0,0) -- (-120:.75cm) arc (-120:-60:.75cm); % mark angles via arc segments
            \draw (-90:0.7cm) node {$60^{\circ}$};
        \end{scope}
        % the triangle
        \draw [line width=1.5pt] (A) -- (B) -- (C) -- cycle;
      \end{tikzpicture}
    \end{figure}
    We have:
    %$$S: (0,+\infty)ightarrow (0,+\infty):\quad S(l)=\frac{\sqrt{3}}{4}a^2.$$
    $$\forall l \in (0,+\infty): \quad S(l)=\frac{\sqrt{3}}{4}a^2.$$

\item[68.] Assume that $w$ is the width of the rectangular box, $h$ is its height. Then the length is $2w.$ Since the volume of the box is $8\mathrm{~ft}^3,$ we have:
    $$2w\cdot w\cdot h=8.$$
    $$2w^2\cdot h=8.$$
    $$h=\frac{8}{2w^2}.$$
    \begin{equation}\label{eq3}h=\frac{4}{w^2}.\end{equation}
    Since the denominator of the fraction cannot be equal to zero, from ef{eq3} we have $w
eq 0$ and $w>0,$ which is equivalent to $w>0.$ Therefore, we have:
    %$$h:(0,+\infty)ightarrow (0,+\infty):\quad h(w)=\frac{4}{w^2}.$$
    $$\forall w \in (0,+\infty):\quad h(w)=\frac{4}{w^2}.$$

\item[69.] Let $l$ be the length of the prism. Then the width is also $l$ because off the square base of the prism. Since the volume of the prism is equal to $2\mathrm{~m}^2,$ we are able to find the height $h$:
    $$h=\frac{2}{l^2}.$$
    Now when we have denoted the values of height, width and length, we can find the surface area of the box:
    $$S=2\left(\frac{2}{l^2}\cdot l + \frac{2}{l^2}\cdot l + l\cdot light);$$
    $$S=2\left(\frac{2}{l} + \frac{2}{l} + l^2ight);$$
    $$S=2\left(\frac{4}{l} + l^2ight);$$
    $$S=\frac{8}{l}+2l^2.$$
    Thus, we can define the function $S$ of $l$ for the surface area as follows:
    % $$S:(0,+\infty)ightarrow (0,+\infty):\quad S(l)=\frac{8}{l}+2l^2.$$
    $$\forall l \in (0,+\infty):\quad S(l)=\frac{8}{l}+2l^2.$$

\item[70.] Let $R$ be the radius of the cylinder and $h$ - its height. 
    \begin{figure}[H]
        \centering
        \begin{tikzpicture}
        % Cylinder: top circle
        \draw[fill=gray, fill opacity=0.1] (0,0) ellipse (1.25 and 0.5);
        % Cylinder: bottom circle
        \draw (-1.25,-3.5) arc (180:360:1.25 and 0.5);
        \draw [dashed] (-1.25,-3.5) arc (180:360:1.25 and -0.5);
        % Cylinder: body
        \draw (-1.25,0) -- (-1.25,-3.5);
        \draw (1.25,-3.5) -- (1.25,0);  
        \fill [gray,opacity=0.1] (-1.25,0) -- (-1.25,-3.5) arc (180:360:1.25 and 0.5) -- (1.25,0) arc (0:180:1.25 and -0.5);
        % Radius
        
ode at (0,0) [circle,fill,inner sep=0.1pt] {};
        \draw [->] (0,0) -- (1.25-0.05,0) node [midway, above, fill=none] {$R$};
        % Height
        
ode at (1.25+0.3,-3.5/2) {$h$};
        \end{tikzpicture}
    \end{figure}
    Therefore, since the volume of the cylinder is equal to $\pi R^2h,$ we have:
    $$\pi R^2h=25.$$
    $$R=\sqrt{\frac{25}{\pi h}}.$$
    Thus, we can define the radius function $R$ of height $h$ as follows:
    %$$R: (0,+\infty)ightarrow (0,+\infty), \quad R(h)=\frac{5}{\sqrt{\pi h}}.$$
    $$\forall h \in (0,+\infty): \quad R(h)=\frac{5}{\sqrt{\pi h}}.$$

\end{enumerate}

Exercise 1.1.65 (65-70)

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Exercise 1.1.65 (65-70)

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