A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is $10$ m. Find the dimensions of the window to admit maximum light through the whole opening.

> [!TLDR] > Eliminating the rectangle's height using the fixed perimeter (10 m) reduces the light-admitting area to a function of the semicircle's radius alone; the area is maximum when $r=h=\dfrac{10}{4+\pi}$ m. ## Setting up variables Let the rectangle have width $2r$ (so the semicircular top has radius $r$) and height $h$. **Area** (rectangle + semicircle on top): $$A=2rh+\frac12\pi r^2$$ **Perimeter:** the window's frame runs along the two vertical sides, the bottom, and the semicircular arc — the top straight edge of the rectangle is replaced by the arc, so it is *not* part of the perimeter. $$P=2h+2r+\pi r=10 \tag{1}$$ > [!WARNING] > A frequent error is to also add the rectangle's top edge to the perimeter — but that edge is where the semicircle joins the rectangle, so it is internal, not part of the outer boundary. ## Reducing to one variable From (1): $h=\dfrac{10-2r-\pi r}{2}$, valid for $0<r<\dfrac{10}{2+\pi}$. Substitute into the area: $$A(r)=2r\cdot\frac{10-2r-\pi r}{2}+\frac12\pi r^2=r(10-2r-\pi r)+\frac{\pi}{2}r^2$$ $$=10r-2r^2-\pi r^2+\frac{\pi}{2}r^2=10r-2r^2-\frac{\pi}{2}r^2$$ ## Differentiating $$\frac{dA}{dr}=10-4r-\pi r=10-r(4+\pi)$$ Set $\dfrac{dA}{dr}=0$: $$r(4+\pi)=10 \implies r=\frac{10}{4+\pi}$$ ## Finding $h$ $$h=\frac{10-2r-\pi r}{2}=\frac{10-r(2+\pi)}{2}$$ …

A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window

Miscellaneous Exercise · Q8

10

Uttar Pradesh UpmspTextbookSubjective· 5mImportance★★★★★

59% · 110/188 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 |

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

Start your 14-day free trial to unlock the full solution →

Eliminating the rectangle's height using the fixed perimeter (10 m) reduces the light-admitting area to a function of the semicircle's radius alone; the area is maximum when $r=h=\dfrac{10}{4+\pi}$ m.

Setting up variables

Let the rectangle have width $2r$ (so the semicircular top has radius $r$ ) and height $h$ .

Area (rectangle + semicircle on top):

A=2rh+\frac12\pi r^2

Perimeter: the window's frame runs along the two vertical sides, the bottom, and the semicircular arc — the top straight edge of the rectangle is replaced by the arc, so it is not part of the perimeter.

P=2h+2r+\pi r=10 \qquad(1)

Watch out

A frequent error is to also add the rectangle's top edge to the perimeter — but that edge is where the semicircle joins the rectangle, so it is internal, not part of the outer boundary.

Reducing to one variable

From (1): $h=\dfrac{10-2r-\pi r}{2}$ , valid for $0<r<\dfrac{10}{2+\pi}$ .

Substitute into the area:

A(r)=2r\cdot\frac{10-2r-\pi r}{2}+\frac12\pi r^2=r(10-2r-\pi r)+\frac{\pi}{2}r^2

=10r-2r^2-\pi r^2+\frac{\pi}{2}r^2=10r-2r^2-\frac{\pi}{2}r^2

Differentiating

\frac{dA}{dr}=10-4r-\pi r=10-r(4+\pi)

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

r(4+\pi)=10 \implies r=\frac{10}{4+\pi}

Finding $h$

$h=\frac{10-2r-\pi r}{2}=\frac{10-r(2+\pi)}{2}$ …

Start your 14-day free trial to unlock the full solution →

←Previous question · misc-exerciseQ7 — The sum of the perimeter of a circle and square is k, where k is some …→Next question · misc-exerciseQ9 — A point on the hypotenuse of a triangle is at distance a and b from th…

Q.A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is $10$ m. Find the dimensions of the window to admit maximum light through the whole opening.

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 $h$

Method: Optimising a Composite-Shape Area Under a Fixed-Perimeter Constraint

Steps

Common Mistakes

Q.A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 101010 m. Find the dimensions of the window to admit maximum light through the whole opening.

Setting up variables

Reducing to one variable

Differentiating

Finding hhh

🎓Unlock everything free for 14 days

Q.A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is $10$ m. Find the dimensions of the window to admit maximum light through the whole opening.

Finding $h$

Unlock everything free for 14 days