Question

Finally, find the dimensions of the cuboid, writing all three on one line.

Answer 1

`\bf \textit{volume of a cuboid}\\\\ V=Lwh~~ \begin{cases} L=length\\ w=width\\ h=height\\[-0.5em] \hrulefill\\ L=4\\ w=x+1\\ h=x+11\\ V=924 \end{cases}\implies 924=4(x+1)(x+11) \\\\\\ \cfrac{924}{4}=(x+1)(x+11)\implies 231=\stackrel{FOIL}{x^2+12x+11} \\\\\\ 0=x^2+12x-220~~ \begin{cases} (-10)(22)=\stackrel{constant}{-220}\\[0.8em] -10+22=\stackrel{\stackrel{middle}{coefficient}}{12} \end{cases} \\\\\\ 0=(x-10)(x+22)\implies x= \begin{cases} \boxed{10}\\ -22 \end{cases}`

it cannot be -22, because it's a dimension and it has to be positive for the cuboid to be there at all.

`\bf \stackrel{length}{4}~\hspace{5em}\stackrel{width}{10+1\implies 11}~\hspace{5em}\stackrel{height}{10+11\implies 21}`