Answer :
Using the joint variation, A similar beam can support [tex]67 \frac{1}{2}[/tex] feet if the beam is one-fourth foot wide and half a foot high, and 12 feet long.
What is Proportion
A proportion is an equation stating that two rational expressions are equal. Simple proportions can be solved by applying the cross products rule.
If [tex]\frac{a}{b} = \frac{c}{d}[/tex] then ab = bc.
What is Direct variation?
The phrase “ y varies directly as x” or “ y is directly proportional to x” means that as x gets bigger, so does y, and as x gets smaller, so does y. That concept can be translated in two ways.
[tex]\frac{y}{x} = k[/tex] for some constant k.
The k is called the constant of proportionality. This translation is used when the constant is the desired result.
What is Inverse Proportion?
According to the expressions "y varies inversely as x" and "y is inversely proportionate to x," y decreases as x increases or vice versa. There are two translations for this idea. For any constant k, referred to as the constant of proportionality, yx = k. If the constant is wanted, use this translation.
What is Joint Variation?
Joint variation is the term used to describe when one variable changes as the sum of other variables. There are two translations for the phrase "y fluctuates concurrently as x and z."
So, In the given question:
Weight that a rectangular beam can support varies jointly as its width and the square of its height and inversely as its length;
It is in joint proportion;
[tex]\begin{aligned}&\mathbf{W}=\frac{\mathbf{w h}^2}{\mathbf{L}} \\&\left\{\begin{array}{l}w=\frac{1}{4} \\h=\frac{1}{3} \\L=12 \\W=30\end{array} \quad \Longrightarrow 30=\frac{\mathbf{k} \frac{1}{4}\left(\frac{1}{3}\right)^2}{12}\right.\end{aligned}[/tex]
[tex]360=\frac{\mathrm{k}}{36} \Longrightarrow 12960=\mathrm{k}[/tex]
Now,
[tex]\text { when }\left\{\begin{array}{l}w=\frac{1}{4} \\h=\frac{1}{2} \\L=12\end{array}\right.[/tex]
[tex]W=\frac{12960\left(\frac{1}{4}\right)\left(\frac{1}{2}\right)^2}{12}[/tex]
[tex]W=1080\left(\frac{1}{4}\right)\left(\frac{1}{4}\right) \\\Longrightarrow W=1080 \cdot \frac{1}{16}[/tex]
[tex]\\\Longrightarrow\mathrm{W}=\frac{135}{2} \\\Longrightarrow \mathrm{W}=67 \frac{1}{2}[/tex]
Hence, A similar beam can support [tex]67 \frac{1}{2}[/tex] feet if the beam is one-fourth foot wide and a half a foot high, and 12 feet long.
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[tex]\bf \qquad \qquad \textit{joint compound proportional variation} \\\\ \begin{array}{llll} \textit{\underline{y} varies directly with \underline{x}}\\ \textit{and inversely with \underline{z}} \end{array}\implies y=\cfrac{kx}{z}\impliedby \begin{array}{llll} k=constant\ of\\ \qquad variation \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{W varies jointly as its width, squared heght and inversely with length}}{W=\cfrac{wh^2}{L}}[/tex]
[tex]\bf \textit{now we also know that }~~ \begin{cases} w=\frac{1}{4}\\ h=\frac{1}{3}\\ L=12\\ W=30 \end{cases}\implies 30=\cfrac{~~k\frac{1}{4}\left( \frac{1}{3} \right)^2~~}{12} \\\\\\ 360=\cfrac{k}{36}\implies 12960=k~\hfill \boxed{W=\cfrac{12960wh^2}{L}}[/tex]
[tex]\bf \textit{when } \begin{cases} w=\frac{1}{4}\\ h=\frac{1}{2}\\ L=12 \end{cases}\textit{ what is \underline{W}?}\qquad W=\cfrac{12960\left( \frac{1}{4} \right)\left(\frac{1}{2} \right)^2}{12} \\\\\\ W=1080\left( \cfrac{1}{4} \right)\left( \cfrac{1}{4} \right)\implies W=1080\cdot \cfrac{1}{16}\implies W=\cfrac{135}{2}\implies W=67\frac{1}{2}[/tex]
