Maximizing a Shop Sign with a Semicircular Top
A shop in Kumamoto wants a new sign whose outline is a rectangle with a semicircle sitting on top; the semicircle’s diameter is exactly the rectangle’s top edge. The outer edge-the two vertical sides, the bottom edge, and the semicircular arc-will be made from a single 6.00 m length of aluminium trim. The diameter itself is an internal joint and uses no trim. The shop owner insists that the sign’s width and height must both be at least 0.50 m, and the trim must be used exactly, with no waste. A designer denotes the semicircle radius by r metres and the rectangle height by h metres, then claims that the area can be treated as a function of r alone. Before cutting metal, the shop wants to know the dimensions that give the largest possible painted area and whether a taller, narrower design would be noticeably worse.
1. Express the painted area A in terms of r only, state the admissible domain for r under the owner's constraints, and explain why h is not an independent variable.
Read the worked answer
The trim length gives the constraint
[2h+2r+\pi r=6.00,]
because the two vertical sides contribute (2h), the base contributes (2r), and the semicircular arc contributes (\pi r). Hence
[h=\frac{6-2r-\pi r}{2}.]
The area is the rectangle plus the semicircle:
[A=2rh+\frac12\pi r^2 = r(6-2r-\pi r)+\frac12\pi r^2 =6r-2r^2-\frac12\pi r^2.]
The width constraint is (2r\ge0.50), so (r\ge0.25). The height constraint is (h\ge0.50), which gives
[\frac{6-(2+\pi)r}{2}\ge0.50 \Rightarrow r\le\frac{5}{2+\pi}\approx0.972.]
Thus (0.25\le r\le0.972). Once the trim length is fixed, choosing (r) forces (h); the two dimensions cannot be varied independently.
2. Find the dimensions that maximize A. Show the derivative and verify that the critical point is a maximum.
Read the worked answer
Differentiate:
[A'(r)=6-4r-\pi r=6-(4+\pi)r.]
Setting (A'(r)=0) gives
[r=\frac{6}{4+\pi}\approx0.840\ \text{m}.]
Then
[h=\frac{6-(2+\pi)(0.840)}{2}\approx0.840\ \text{m}.]
Since (A''(r)=-(4+\pi)<0), this critical point is a maximum. The maximum area is approximately
[A=6(0.840)-2(0.840)^2-\frac12\pi(0.840)^2\approx2.52\ \text{m}^2.]
The endpoints give smaller areas: at (r=0.25), (A\approx1.28\ \text{m}^2); at (r\approx0.972), (A\approx2.46\ \text{m}^2). So the optimal sign is almost square in its rectangular part, with (r\approx h\approx0.84\ \text{m}).
3. The owner proposes fixing h = 1.00 m to make the sign taller. Using your area function, judge whether this increases or decreases the painted area, and quantify the loss.
Read the worked answer
With (h=1.00), the trim constraint gives
[2(1.00)+(2+\pi)r=6.00 \Rightarrow r=\frac{4}{2+\pi}\approx0.778\ \text{m}.]
The area is then
[A=2(0.778)(1.00)+\frac12\pi(0.778)^2 \approx1.556+0.950=2.506\ \text{m}^2.]
This is about (0.014\ \text{m}^2) less than the maximum (2.520\ \text{m}^2), a loss of roughly 0.6%. So the taller design is not meaningfully worse in painted area; the choice can be decided by legibility and appearance rather than by area alone. The small loss occurs because the maximum is flat near (r\approx0.84), so modest changes in shape cost very little area.