Resources for Applications of Maxima and Minima
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Questions
19
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Video Tutorials
2
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HSC Questions
10
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Applications of Maxima and Minima Theory
![There are two types of applications of maxima and minima questions. For the easier question you are given a function that requires you to determine the maximum or minimum value. \\ The harder question requires you to derive a function which then requires you to determine the maximum or minimum value.\\ \textbf{Example}\\ An open rectangular box has four sides and a base, but no lid, as in the figure.\\ \begin{center} \begin{tikzpicture} \def\l{3} \def\w{1} \def\h{1.2} \coordinate (A) at (0,0,0); \coordinate (B) at (\l,0,0) ; \coordinate (C) at (\l,\w,0); \coordinate (D) at (0,\w,0); \coordinate (E) at (0,0,\h); \coordinate (F) at (\l,0,\h); \coordinate (G) at (\l,\w,\h); \coordinate (H) at (0,\w,\h); %Ecken \node[left= 1pt of A]{}; \node[right= 1pt of B]{}; \node[right= 1pt of C]{}; \node[left= 1pt of D]{}; \node[left= 1pt of E]{}; \node[right= 1pt of F]{}; \node[right= 1pt of G]{}; \node[left= 1pt of H]{}; %Kanten \draw[] (A) -- (B) node[midway, below]{} -- (C) node[midway, right]{} -- (D) node[midway, above]{} -- (A) node[midway, left]{}; \draw[] (B) -- (F) -- (G) -- (C); \draw[] (G) -- (H) -- (D); \draw[] (A) -- (E)-- (F) node[midway,below]{\(\tiny 3x\)}; \draw[] (E) -- (H); \path (F)--(B) node[midway,below right]{\(\tiny x\)}; \path (B)--(C) node[midway,right]{\(\tiny y\)}; \path[draw,fill=gray!10!white] (E)--(F)--(G)--(H)--cycle; \path[draw,fill=gray!10!white] (F)--(B)--(C)--(G)--cycle; \end{tikzpicture} %\includegraphics[width=0.3\textwidth]{72542003-46c9-41b7-b8be-06d7096b0c23} \end{center} The dimensions of the base are: length \(3x \) cm, width \(x \) cm and height \(y\) cm \begin{itemize} \item[\bf{i)}]Given that the surface area of the box is \(120c{m^2}\) show that \(y = \dfrac{{120 - 3{x^2}}}{{8x}}\) \item[\bf{ii)}]Show that the volume of the box is \(V = 9x\left( {5 - \dfrac{{{x^2}}}{8}} \right)\) \item[\bf{iii)}]Hence determine the maximum volume of the box \end{itemize} \vfill \begin{multicols}{2} \textbf{Solution}\\ $\begin{aligned} \text{(i)}\quad SA &=3 x^{2}+2 x y+6 x y \\ &=3 x^{2}+8 x y \\ \therefore 3 x^{2} &+8 x y=120 \\ & 8 x y=120-3 x^{2} \\ & \therefore y=\frac{120-3 x^{2}}{8 x} \end{aligned}$\\ $\begin{aligned}\text{(ii)}\quad V &=3 x \times x \times y \\ &=3 x^{2} \times \frac{120-3 x^{2}}{8 x} \\ &=\frac{3 x}{8} \times 3\left(40-x^{2}\right) \\ &=\frac{9 x}{8}\left(40-x^{2}\right)\\ &=9 x\left(5-\frac{x^{2}}{8}\right) \end{aligned}$\\ \columnbreak $\begin{aligned}\text{(iii)}\quad V &=45 x-\frac{9 x^{3}}{8} \\ V^{\prime} &=45-\frac{27 x^{2}}{8} \\ V^{\prime \prime} &=-\frac{54 x}{8} \rightarrow \max (x>0) \\ \text { Let}\ V^{\prime} &=0 \quad x^{2}=\frac{45 \times 8}{27} \rightarrow x=\sqrt{\frac{40}{3}} \\ y &=\frac{120-3 \times \frac{40}{3}}{8 \sqrt{\frac{40}{3}}} \\ &=10 \times \sqrt{\frac{3}{40}} \\ V &=3 \times \frac{40}{3} \times 10 \sqrt{\frac{3}{40}} \\ &=20 \sqrt{30}\mathrm{~cm}^{3} \end{aligned}$\\ \end{multicols}](/media/olxl0gkr/4867.png)