An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a giv

Question 187 of 188

Q. An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of material will be least when depth of the tank is half of its width. If the cost is to be borne by nearby settled lower income families, for whom water will be provided, what kind of value is hidden in this question?

Odisha ChseCBSE Class XII Board 2018Subjective· 4mImportance★★★★★

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Concept understanding — Optimization Word Problem

Optimization Word Problems

Imagine planning a garden with 40 metres of fencing and wanting the largest rectangular area. A long, thin rectangle wastes space; a square feels roomier; somewhere in between lies the best shape. That is an optimisation problem — a fixed resource and a quantity to make as large (or as small) as possible.

Every optimisation word problem has the same skeleton: the best outcome — maximum area, minimum cost, largest volume, shortest time — under a constraint — limited material, a fixed budget, a given perimeter.

The Plan of Attack

The problem gives you a story, not a graph. Your job is to turn it into a single-variable function and find its peak or valley:

Name the quantity to optimise — call it $Q$ , and write it using variables.
Find the constraint — a relation between those variables (e.g. "perimeter $=40$ ").
Reduce to one variable — use the constraint to eliminate the rest.
Differentiate — solve $Q'(x)=0$ to find the critical points.
Confirm — use $Q''(x)<0$ for a maximum or $Q''(x)>0$ for a minimum.
Answer the question asked — give the actual dimensions/cost, not just $x$ .

Note

In board exams these problems almost always reduce to a quadratic or cubic. Once $Q(x)$ is written, the calculus is mechanical.

The Garden, Worked

40 m of fencing encloses a rectangle; maximise the area.

Objective: $A = l\,w$ .
Constraint: $2l+2w=40$ , so $l+w=20$ .
Reduce: $w=20-l$ , giving $A(l)=l(20-l)=20l-l^2$ .
Differentiate: $A'(l)=20-2l=0 \implies l=10$ .
Confirm: $A''(l)=-2<0$ , a maximum.

So $l=w=10$ m — a 10 m $\times$ 10 m square.

Watch out

A common slip: solving $A'(l)=0$ and stopping. Always check max vs min, and answer in the units asked.

The Common Families

| Problem type | Typical objective | Typical constraint |

|--------------|-------------------|--------------------| …

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Minimum-cost tank has depth equal to half its width; the hidden value is care and social responsibility toward lower-income families.

Concept. Minimise a function subject to a fixed-volume constraint using the first and second derivative tests.

Why this method. The material used equals the surface area of the open tank; expressing it in one variable and minimising gives the optimum shape.

Working. Let the square base have side $x$ and depth $h$ , with fixed volume $V=x^2h$ , so $h=\dfrac{V}{x^2}$ . For an open tank (no top), the material (cost $\propto$ area) is

$S=\underbrace{x^2}_{\text{base}}+\underbrace{4xh}_{\text{4 sides}}=x^2+4x\cdot\frac{V}{x^2}=x^2+\frac{4V}{x}.$

$\frac{dS}{dx}=2x-\frac{4V}{x^2}=0\ \Rightarrow\ 2x^3=4V\ \Rightarrow\ x^3=2V.$ …

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