Q.119. Maxima

Question 119.

For the 2 unit squares given in the drawing below, find the exact value of the maximal height of the vertical line segment AB. You may express it by using φ, the golden ratio.

Answer 119.

If θ = angle made by the side of the oblique square with the horizontal, then taking its vertex on the horizontal as origin, the coordinates of the two vertices, C and D are

(1 + cosθ + sinθ,1) and
[ sinθ + √2 cos(45° + θ), √2 sin(45° + θ) ] = [ cosθ, (cosθ + sinθ) ]

Slope of the line, m = (1 - cosθ - sinθ) / (1 + sinθ)

Its equation as it passes through (1 + cosθ + sinθ, 1) is
y - 1 = m [x - (1 + cosθ + sinθ)]
=> AB
= 1 - m(1 + cosθ + sinθ)
= 1 - (1 + cosθ + sinθ) (1 - cosθ - sinθ) /(1+ sinθ)
= 1 + 2sinθcosθ/(1 + sinθ)

For AB to be maximum, d/dθ (AB) = 0
=> (1 + sinθ)(2cos^2 θ - 2sin^2 θ) - 2sinθ cos^2 θ = 0
=> 2(cos^2 θ - sin^2 θ - sin^3 θ) = 0
=> sin^3 θ + 2sin^2 θ - 1 = 0
=> sin^3 θ + sin^2 θ + sin^2 θ - 1 = 0
=> sin^2 θ (sinθ + 1) + (sinθ +1)(sinθ - 1) = 0
=>(sinθ + 1) (sin^2 θ + sinθ - 1) = 0
sinθ + 1 = 0 is obviously ruled out

=> sin^2 θ + sinθ - 1 = 0
=> sinθ = (1/2) [√5 - 1] = φ - 1
=> 1 + sinθ = φ
=> sinθ = (1/2) (√5 - 1) * [(√5 + 1)/(√5 + 1)] = 1/φ
and cosθ = √(1 - 1/φ^2)=1/√φ

=> AB
= 1 + 2sinθ cosθ /(1 + sinθ)
= 1 + 2*[(1/φ)*(1/√φ)]/φ
= 1 + 2(φ)^(-5/2)   ...     [Exact Value]
= 1.600566   ...   [Approximate Value].

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