Miscellaneous · Q6: Open tank, depth 2 m, volume 8 m³: least cost Rs 1000

Basics you need

Solution

1. Variables, and what the volume tells us

   Let the base be $l$ m long and $w$ m wide. The depth is 2 m, so

   $$\text{volume} = 2lw = 8 \quad\Longrightarrow\quad lw = 4, \quad w = \frac{4}{l}.$$

   Range. Any $l > 0$ gives a genuine tank (with $w = \frac4l > 0$⁠), so the allowed range is $l > 0$⁠.

   Two unknowns, one condition: the volume links them, so we can write $w$ in terms of $l$ and work with one variable.
   Volume of a cuboid (Basic 2). Basics: area and volume formulas

2. Areas of the base and the sides

   Base: $lw = 4$ m², whatever $l$ is. Sides: two of area $2l$ and two of area $2w$⁠, in total

   $$2(2l) + 2(2w) = 4(l + w) \text{ m}^2.$$
   An open tank has a base and four sides but no lid, so only these five faces are paid for. The base area is the same for every allowed tank, so the base cost is a fixed amount.

3. The cost as a function of $l$

   $$\begin{aligned} C &= 70 \times 4 + 45 \times 4(l + w) \\ &= 280 + 180\left(l + \frac4l\right), \quad l > 0. \end{aligned}$$
   Cost = rate $\times$ area, for the base and for the sides separately, then added. Replacing $w$ by $\frac4l$ leaves a function of $l$ alone.
   Basics: building the function to optimise

4. Differentiate and find the critical point

   $$\begin{aligned} C'(l) &= 180\left(1 - \frac{4}{l^2}\right) \\ &= \frac{180(l^2 - 4)}{l^2} \\ &= \frac{180(l - 2)(l + 2)}{l^2}. \end{aligned}$$

   $C'(l) = 0$ at $l = 2$ or $l = -2$⁠; only $l = 2$ is in the range $l > 0$⁠.

   The constant 280 has derivative 0, and $\frac{d}{dl}\frac4l = -\frac{4}{l^2}$⁠. Factorising $l^2 - 4$ as a difference of squares shows the sign of $C'$ at a glance.
   Power rule (Basic 3).

5. $l = 2$ gives the least cost on the whole range

   For $l > 0$⁠: $l + 2 > 0$ and $l^2 > 0$⁠, so $C'(l)$ has the sign of $l - 2$⁠:

   • $0 < l < 2$⁠: $C'(l) < 0$⁠, so $C$ is decreasing on $(0, 2]$⁠;
   • $l > 2$⁠: $C'(l) > 0$⁠, so $C$ is increasing on $[2, \infty)$⁠.

   Hence $C(l) > C(2)$ for every $l \ne 2$ (Theorem 1 and its proof). As a second check, $C''(l) = \frac{1440}{l^3}$⁠, so $C''(2) = 180 > 0$⁠, a local minimum by Theorem 4 (ii), in agreement.

   The range $l > 0$ has no endpoints to test, so the sign of $C'$ on the whole range does all the work: very thin tanks (small $l$⁠) and very long tanks (large $l$⁠) both cost more, because their sides have a large total area.
   Theorem 1, Section 6.3; Theorem 4, Section 6.4 (Basic 4). Lesson: optimisation problems

6. The least cost

   At $l = 2$⁠: $w = \frac42 = 2$⁠, and

   $$C(2) = 280 + 180(2 + 2) = 280 + 720 = 1000.$$

   The cheapest tank has a 2 m by 2 m base (so it is a 2 m cube without a lid), and it costs Rs 1000.

Every fact used, and where it comes from