Answer:
To minimize cost, the pipe should follow a rectangular path.
Explanation:
Let us denote the square bottom of the dug pit as
P = S²
and its cost will be P = 2S² = 4S to dig
Let us denote the height of the pit as H,
Therefore, Total area of the pit will be:
4SH
and its cost is 2SH
The volume of the pit is
v= S²H = 4² × 2 = 128
Total cost therefore is =
C = 2S² + 2SH
This therefore translates to
128 = s²h
Making H the subject of the formula,
we have, h = 128 / S²
Cs therefore = 2S² +2S . 128 / S²
Cs = 2S² + 256 / S
Cs = 4S - 256 / S²
Cs = 4S³ - 256 / S
4S³ - 256 / S = 0
4S³ - 256 - 0 = S
Solving further, S = 4
Where,
H = 128/64
= 2
Therefor, S = 4 and H= 2
So our pipe will follow a rectangular path to properly minimize cost.