The sum of the perimeter of a circle and square is $k$, where $k$ is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

> [!TLDR] > Eliminating the square's side using the fixed-perimeter constraint reduces the total area to a function of the circle's radius alone; differentiating shows the area is least exactly when the side of the square equals twice the radius, i.e. $a=2r$. ## Setting up variables Let the circle have radius $r>0$ and the square have side $a>0$. Perimeter of circle $=2\pi r$, perimeter of square $=4a$. Given their sum is fixed: $$2\pi r+4a=k \quad (k\text{ constant}) \tag{1}$$ Sum of areas to be minimised: $$S=\pi r^2+a^2 \tag{2}$$ ## Reducing to one variable From (1): $a=\dfrac{k-2\pi r}{4}$, valid for $0<r<\dfrac{k}{2\pi}$. Substitute into (2): $$S(r)=\pi r^2+\left(\frac{k-2\pi r}{4}\right)^2=\pi r^2+\frac{(k-2\pi r)^2}{16}$$ ## Differentiating $$\frac{dS}{dr}=2\pi r+\frac{2(k-2\pi r)}{16}\cdot(-2\pi)=2\pi r-\frac{\pi(k-2\pi r)}{4}$$ Set $\dfrac{dS}{dr}=0$: $$2\pi r=\frac{\pi(k-2\pi r)}{4} \implies 8r=k-2\pi r \implies r(8+2\pi)=k \implies r=\frac{k}{2(\pi+4)}$$ ## Finding $a$ and checking the required relation $$a=\frac{k-2\pi r}{4}=\frac{k-2\pi\cdot\frac{k}{2(\pi+4)}}{4}=\frac{k\left(1-\dfrac{\pi}{\pi+4}\right)}{4}=\frac{k\cdot\dfrac{4}{\pi+4}}{4}=\frac{k}{\pi+4}$$ Compare with $2r$: …

The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their ar

Miscellaneous Exercise · Q7

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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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Eliminating the square's side using the fixed-perimeter constraint reduces the total area to a function of the circle's radius alone; differentiating shows the area is least exactly when the side of the square equals twice the radius, i.e. $a=2r$ .

Setting up variables

Let the circle have radius $r>0$ and the square have side $a>0$ .

Perimeter of circle $=2\pi r$ , perimeter of square $=4a$ . Given their sum is fixed:

2\pi r+4a=k \quad (k\text{ constant}) \qquad(1)

Sum of areas to be minimised:

S=\pi r^2+a^2 \qquad(2)

Reducing to one variable

From (1): $a=\dfrac{k-2\pi r}{4}$ , valid for $0<r<\dfrac{k}{2\pi}$ .

Substitute into (2):

S(r)=\pi r^2+\left(\frac{k-2\pi r}{4}\right)^2=\pi r^2+\frac{(k-2\pi r)^2}{16}

Differentiating

\frac{dS}{dr}=2\pi r+\frac{2(k-2\pi r)}{16}\cdot(-2\pi)=2\pi r-\frac{\pi(k-2\pi r)}{4}

Set $\dfrac{dS}{dr}=0$ :

2\pi r=\frac{\pi(k-2\pi r)}{4} \implies 8r=k-2\pi r \implies r(8+2\pi)=k \implies r=\frac{k}{2(\pi+4)}

Finding $a$ and checking the required relation

a=\frac{k-2\pi r}{4}=\frac{k-2\pi\cdot\frac{k}{2(\pi+4)}}{4}=\frac{k\left(1-\dfrac{\pi}{\pi+4}\right)}{4}=\frac{k\cdot\dfrac{4}{\pi+4}}{4}=\frac{k}{\pi+4}

Compare with $2r$ : …

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Q.The sum of the perimeter of a circle and square is $k$ , where $k$ is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

Concept understanding — Optimization Word Problem

Optimization Word Problems

The Plan of Attack

The Garden, Worked

The Common Families

Setting up variables

Reducing to one variable

Differentiating

Finding $a$ and checking the required relation

Method: Lagrange Multipliers for a "Prove the Minimum Occurs When..." Problem

Steps

Common Mistakes

Q.The sum of the perimeter of a circle and square is kkk, where kkk is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

Setting up variables

Reducing to one variable

Differentiating

Finding aaa and checking the required relation

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Q.The sum of the perimeter of a circle and square is $k$ , where $k$ is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

Finding $a$ and checking the required relation

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