Largest Rectangle in a Parabolic Arch
A designer is fitting a rectangular glass panel inside a parabolic arch whose inner edge is modelled by y = 9 − x² for −3 ≤ x ≤ 3, with x and y measured in metres and the ground at y = 0. The panel's base rests on the ground and its two upper corners touch the arch, so the panel is symmetric about the y-axis. A decorative metal strip is to be fixed along the tangent to the arch at the point where x = 1. The designer says the largest possible panel is the tallest one that fits, and also wants to know what fraction of the arch opening the largest panel would cover so that glazing and frame can be costed. Frame width is taken as negligible.
1. Find the dimensions and area of the largest rectangular panel that fits, and assess the designer's claim that the tallest panel is the largest.
Read the worked answer
Place the base corners at x = −a and x = +a, so the width is 2a and the height is the arch height there, y = 9 − a². Then A(a) = 2a(9 − a²) = 18a − 2a³, with 0 < a < 3.
Differentiate: A′(a) = 18 − 6a². Setting A′ = 0 gives a² = 3, so a = √3 (the negative root is outside the interval). Justify the maximum: A″(a) = −12a, so A″(√3) = −20.8 < 0 - concave down, a local maximum. Since a = √3 is the only critical point in (0, 3) and the endpoints give A(0) = 0 and A(3) = 0, it is the global maximum.
Dimensions: width = 2√3 ≈ 3.46 m; height = 9 − 3 = 6.0 m; area = 6 × 2√3 = 12√3 ≈ 20.8 m².
The claim is false. The tallest panel is the one with a → 0, height 9 m, but its width tends to 0 and its area to 0. Tallest and largest are different objectives: area is width × height, and widening the panel is worth more than the height lost while a < √3.
2. Find the equation of the tangent to the arch at x = 1, and show that the line meets the curve only at that point.
Read the worked answer
For y = 9 − x², dy/dx = −2x. At x = 1 the slope is −2 and the point of contact is (1, 9 − 1) = (1, 8).
Tangent: y − 8 = −2(x − 1) → y = 10 − 2x.
Intersection with the parabola: 9 − x² = 10 − 2x → x² − 2x + 1 = 0 → (x − 1)² = 0. This has the repeated root x = 1 (discriminant 4 − 4 = 0), so the line touches the curve at exactly one point and does not cross it - the definition of tangency.
Side check: y_line − y_curve = (10 − 2x) − (9 − x²) = x² − 2x + 1 = (x − 1)² ≥ 0 for all x, so the line lies on or above the arch and touches only at x = 1. The strip therefore sits outside the glazed opening, which is why the designer can fix it without reducing the panel.
3. Calculate the area of the arch opening and the fraction of it covered by the largest panel.
Read the worked answer
Area under the arch (between the curve and the ground, x = −3 to 3): A = ∫₋₃³ (9 − x²) dx = [9x − x³/3]₋₃³ = (27 − 9) − (−27 + 9) = 18 − (−18) = 36 m². The integrand is even, so this equals 2 × ∫₀³ (9 − x²) dx = 2 × (27 − 9) = 36 m², a useful check.
Fraction covered by the largest panel: 12√3 / 36 = √3/3 ≈ 1.732/3 = 0.577, about 58%.
So just over half the opening is glazed, leaving about 42% (≈15.2 m²) of the parabolic area as margin around the panel. Note that the panel's area is not simply the fraction of the width times the fraction of the height - the arch shape means the corners cut away a substantial share, which is exactly what the integration quantifies.