Question

Pool-O is a pool cleaning robot. When it's cleaning a pool, it first drains the pool, then scrubs the walls and the floors of the pool, and finally refills the pool.
Pool-O pumps 5 cubic meters per minute (both when draining and when filling) and scrubs 8 square meters per minute.
Write an equation that represents the relationship between the pool's width (w), the pool's length (l), the pool's depth (d), and the number of minutes it takes Pool-O to clean it (t).
All lengths are in meters. This question is very confusing someone plz explain this.

Answer 1

\[T = 0.4 * w * l *d + w*l + 2w *d +2l *d/8\] Someone plz tell me how I got this answer. I don't understand.

Answer 2

" pool's width (w), the pool's length (l), the pool's depth (d) "

What is the total volume of water in the pool?

Answer 3

Volume of the pool:
\[V~\text{m}^3=(L~\text{m})\times (W~\text{m})\times (D~\text{m})\]
When pumping out, \(V\) changes by \(\dfrac{5\text{m}^3}{\text{min}}\):
\[\frac{5~\text{m}^3}{\text{min}}=\frac{(L~\text{m})\times (W~\text{m})\times (D~\text{m})}{\text{min}}\]
You want to be able to describe the time it takes for the pool to empty, so you try to express time in terms of the volume and rate at which the volume is taken out:
\[\begin{align*}\text{min}&=\frac{(L~\text{m})\times (W~\text{m})\times (D~\text{m})}{\dfrac{5~\text{m}^3}{\text{min}}}\\\\
T&=\frac{L\times W\times D}{5}\text{min}
\end{align*}\]
The same rate is used for refilling the pool, so you essentially double the time it takes:
\[T=\frac{2(L\times W\times D)}{5}\text{min}=0.4(L\times W\times D)\]
Surface area of the pool:
\[A~\text{m}^2=[(L~\text{m})\times (W~\text{m})]+2[(W~\text{m})\times (D~\text{m})]+2[(L~\text{m})\times (D~\text{m})]\]
When scrubbing, \(A\) changes by \(\dfrac{8~\text{m}^2}{\text{min}}\):
\[\begin{align*}\frac{8~\text{m}^2}{\text{min}}&=\frac{[(L~\text{m})\times (W~\text{m})]+2[(W~\text{m})\times (D~\text{m})]+2[(L~\text{m})\times (D~\text{m})]}{\text{min}}\\\\
\text{min}&=\frac{[(L~\text{m})\times (W~\text{m})]+2[(W~\text{m})\times (D~\text{m})]+2[(L~\text{m})\times (D~\text{m})]}{\dfrac{8~\text{m}^2}{\text{min}}}\\\\

\text{min}&=\frac{[(L\times W)~\text{m}^2]+2[(W\times D)~\text{m}^2]+2[(L\times D)~\text{m}^2]}{\dfrac{8~\text{m}^2}{\text{min}}}\\\\
T&=\frac{(L\times W)+2(W\times D)+2(L\times D)}{8}\text{min}
\end{align*}\]
The total time would then be the sum of these two \(T\) expressions, precisely what you're supposed to get.

Answer 4

thx for helping me.