Optimizing volume with an irregular shape: A 5th Grade Math Problem Misinterpreted

Question

I was helping my daughter with math tonight when I initially misinterpreted the problem below to have to come up with what the values for length, width, and depth are when the total volume is exactly equal to 300 and is comprised of 6 equal cubes.

This left me wondering if the question had been as I initially interpreted, what would the algebraic approach have been?

I brute forced finding an answer with a function in R but I'd be interested to know if there was a simpler way to solve for reasonable (within 3 decimals) values of the three dimensions?

To find reasonable values, I approached it as if it were a rectangular prism (volume = 400) and divided it into 8 equal cubes (each with a volume = 50) and then cut out two cubes for a volume of 300.

R code below

library(dplyr)
expand.grid(length=seq(1,4,0.1), width=seq(10,16,0.1)/4, height=seq(6,8,0.1)/2) %>% 
  mutate(prod=apply(X=.,MARGIN=1,FUN=prod)) %>% 
  filter(prod==50)

results

    length   width    height   prod
1   4        3.125    4        50

My initial answer: length=12 width=8 depth=3.125

Answer 1

Let $L$ denote the total length, $W$ width, and $D$ the deepest part of the pool, $V$ it's volume. Additionally we need to know $d$=the depth of the shallow part of the pool, and $l$ the length of each shallow section of the pool. With this notation we have $0\leq V = WD(L-2l) + 2Wdl \leq 300$. From the figure we require that $0<d<D$ and $0<l<L/2$.

Note that we do not have a (nonzero) lower bound on the volume, so it is beneficial to take the dimensions to be as small as possible. Furthermore, there is no (nonzero) lower bound on the depth $D$, so we can always make sure that the volume is within bounds by making the pool as shallow as we like.

With this in mind, let us get a feeling for the numbers by taking $L=10$, $W=6$, $D=4$, $l=0$, $d=D$, so that the pool is just a box with the volume $V=WDL=240$ cubic feet, which is already within the specs. Therefore any $0<l<5$ and $0<d<4$ will solve the problem, e.g. $l=2$ and $d = 2$ will do.

For any other valid choices of $L$ and $W$ one can always start by computing $D$ which satisfies or almost satisfies the volume requirement with $d=D$ and $l=0$, and then cut off excess volume by increasing $l$ and decreasing $D$.