Minimum-Material Dimensions of a Fixed-Volume Open-Top Tank: 23-Chem-B3 Simulation, Modelling, and Optimization, December 2014, Q1

Question 1: Minimum-Material Dimensions of a Fixed-Volume Open-Top Tank (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. An open-top rectangular tank of fixed volume $V = 500\ \text{{m}}^3$, with base $x \times y$ and depth $h$. Because the bottom and the four side walls have the same thickness, the material (cost) is proportional to the exposed surface area — bottom plus four walls, with no lid.

Find. The dimensions $x,\,y,\,h$ that minimize the surface area (hence the material cost) at the fixed volume.

Figure 1 — Open-top rectangular tank: base x by y, depth h, fixed volume V = x·y·h = 500 m³. Cost is proportional to the bottom + four walls (no lid).

Approach. Write the surface area as the objective and the fixed volume as the equality constraint, then minimize with Lagrange multipliers; the stationarity conditions force a square base and $h = x/2$.

Objective and constraint. With uniform thickness the cost tracks the area of the five plates (no top): $$A = \underbrace{xy}_{\text{bottom}} + \underbrace{2xh + 2yh}_{\text{four walls}}, \qquad \text{subject to}\qquad xyh = V = 500.$$
Stationarity (Lagrange). With $\mathcal{{L}} = xy + 2xh + 2yh - \lambda(xyh - V)$, setting $\partial\mathcal{{L}}/\partial x=\partial\mathcal{{L}}/\partial y=\partial\mathcal{{L}}/\partial h=0$ gives $$y + 2h = \lambda yh,\quad x + 2h = \lambda xh,\quad 2x + 2y = \lambda xy.$$ Subtracting the first two equations yields $(y - x) = \lambda h (y - x)$, so the optimum has $x = y$ (a square base).
Relate depth to base. Putting $x=y$ into the third condition, $4x = \lambda x^2 \Rightarrow \lambda = 4/x$; substituting into the first gives $x + 2h = 4h$, i.e. $$\boxed{h = \tfrac{1}{2}x}\quad(\text{depth is half the base side}).$$
Impose the volume. With $x=y$ and $h=x/2$, $\;V = x^2 h = x^3/2 = 500$, so $$x^3 = 1000 \;\Rightarrow\; x = y = \boxed{10\ \text{{m}}},\qquad h = \boxed{5\ \text{{m}}}.$$
Minimum material. The corresponding plate area is $$A_{{\min}} = x^2 + 4xh = 100 + 4(10)(5) = \boxed{300\ \text{{m}}^2},$$ and the second-order (bordered-Hessian) test — or simply noting $A\to\infty$ as $x\to0$ or $x\to\infty$ — confirms this stationary point is the minimum.

Quantity	Result
Base (square)	$x = y = 10\ \text{{m}}$
Depth	$h = 5\ \text{{m}}$ ($=x/2$)
Minimum plate area (material)	$A_{{\min}} = 300\ \text{{m}}^2$

Concept & Theory

Topic. Constrained optimization of a geometric design by Lagrange multipliers (an equality-constrained NLP).

Key relations. Objective $A = xy + 2h(x+y)$; constraint $g = xyh - V = 0$; stationarity $\nabla A = \lambda \nabla g$. The result $h = x/2$ for a square base is the classic open-top-box optimum.

Why this works. The Lagrange conditions equalize the marginal "area cost" per unit volume in every direction. Symmetry between $x$ and $y$ in both objective and constraint forces $x=y$; the absence of a top halves the effective wall demand in the depth direction, which is why the optimal depth is only half the base side (a closed box would give a cube).

Common pitfall. Including a lid (using $A = 2xy + 2xh + 2yh$) gives a cube and the wrong dimensions. Read whether the tank is open- or closed-top before writing the area.

Source. Edgar, Himmelblau & Lasdon, Optimization of Chemical Processes, ch. 8 (equality-constrained optimization); Rao, Engineering Optimization §2.4 (Lagrange multipliers).

Question 1 of 7: Minimum-Material Dimensions of a Fixed-Volume Open-Top Tank

Question 1: Minimum-Material Dimensions of a Fixed-Volume Open-Top Tank (20 marks)