maximum trapezium in a semicirle.?

Question

I need to find the trapezium with the maximum area which is in a semicircle, using differentiation. The big base of the trapezium is the diameter of the semicircle (2R) and the other two vertices on the semicircle.
What I have to do, is to express the area A=[(b1+b2)h]/2 in R, then find the first derivative and then the stationary points.
Any idea how to express all the variables of the formula in R?

Thank you all for your effort and help.
Since the question was to find this by differentiating the formula of the trapezium, after having R the one and only variable, I would choose smci's solution as best one. My difficulty was to find the way to express A in R.
Thank you.

Answer 1

[I wasn't sure if you wanted an overly general version with two independent variable sidelengths, but Madhukar sent me a clarification that b1,b2 are opposite sides.
b1=2R is the base and b2 is our variable.
Madhukar also kindly found the error in my A'(h)]

So since h is the height, and by symmetry height h bisects unknown side b2 and is orthogonal to it, and the hypotenuse of that triangle is R since the vertex is on the circle:
=> b2=2√(R² - h²)

Thus area A = [(b1+b2)h]/2
= [(2R+2√(R² - h²))h]/2
= (R+√(R² - h²))h
= hR+h√(R² - h²)
A(h) = R² [ (h/R) + (h/R)√(1 - (h/R)²)

If we transform to x = h/R
A(x) = R² [ x + x√(1-x²) ]

Differentiating and using the Product Rule:
A'(x) = R² [ 1 + √(1-x²) + x(-2x)/2√(1-x²) ]
A'(x) = R² [ 1 + √(1-x²) - x²/√(1-x²) ]
A'(x) = R² [ 1 + (1-x² -x²)/√(1-x²) ]
A'(x) = R² [ 1 + (1-2x²)/√(1-x²) ]

A'(x) = 0 when 1 + (1-2x²)/√(1-x²) = 0
(2x²-1)/√(1-x²) = 1
(2x²-1)² = (√(1-x²))²
4x^4 -4x²+1 = 1-x²
4x^4 -3x² = 0
x²(4x²-3) = 0
=> x = +√3 /2 ; [ignoring x = 0 which is a minimum]

So your answer is optimal h = √3R /2
and A_max ≈ R² (√3 /2)(1 + √(1 - 3/4))
= 3√3 /4 R²

[I revised this after Madukar pointed out the algebraic error in A'(h)]

[I was expecting the Cartesian formulation would be more elegant than polar, but polar is more elegant.
And Scythian's is by far the most elegant proof.]

Answer 2

Given any chord of a circle, the triangle having maximum possible area is an isoceles with the chord as the base. Therefore the trapezium must have three equal sides besides the base of the semicircle, i.e. a semi-hexgaon. QED

Want to find out by differentiation? Find the maximum by differentiating Sin(A/2) + Sin(B/2) + Sin((π-A-B)/2). A and B would be the angles subtended by 2 sides of the trapezium. You'll end up with A = B = π/3 = 60 degrees.

The area of the semi-hexagon in terms of R would be (3/4)√3 R² = 1.299 R², as compared to the area of the semi-circle, which is (π/2) R² = 1.57 R².

Answer 3

If you draw the figure, using pythagoras theorem, you can prove that
b1 = 2√(R^2 - h^2), and obviously b2 = 2R

=> A = (h/2)[ 2√(R^2 - h^2) + 2R ]
=> dA/dh = (h/2) [ -2h / √(R^2 - h^2) ] + (1/2) )[ 2√(R^2 - h^2) + 2R ]
From here, you can proceed equating dA/dh to zero and find h for maximum area.

But, it will better to follow the analytical geometry method.
Let the equation of the circle be x^2 + y^2 = R^2.
Let one upper vertex of trapezium be (Rcos θ, Rsin θ).
Then the other vertex will be (-Rcosθ, Rsin θ)
b1 = 2Rcos θ, b2 = 2R and h = Rsin θ
A = (h/2)(b1 + b2) = [(Rsin θ)/2]*[2Rcos θ + 2R]
=> A = (R^2) * (sin θcos θ + sin θ)
=> dA/dθ = (R^2) * (co^2 θ - sin^2 θ + cos θ)
=> dA/dθ = (2R^2) * (cos 2θ + cos θ)

For A to be maximum, dA/dθ = 0 and d2A/dθ2 < 0.
=> cos 2θ = - cos θ = cos (π - θ)
=> π - θ = 2kπ ± 2θ, k ∈ Z
=> 3θ = π - 2kπ or θ = 2kπ - π k ∈ Z
=> θ = π/3 - 2kπ/3 or θ = 2kπ - π k ∈ Z
Only plausible value of θ = π/3
It can be proved that d2A/dt2 < 0 for θ = π/3
Max. area
A = (R^2) * (sin π/3 cos π/3 + sin π/3)
= [(3√3) /4] R^2.
= 1.3 R^2 (nearly).