Calculus question about rectangle inscribed in a semi circle?

Question

A rectangle is to be inscribed in a semicircle of radius r with its base touching that of the semicircle. What are the dimensions of rectangle if its area is to be maximized?

Dimensions:_____________r X _______________r

Does anyone have any idea how to do this problem. I really think my calc professor has it out for his students and these problems seem ridiculous. Thanks for any help provided.

Answer 1

Let dimensions of rectangle be a and 2b (b either side of the centre on the diameter)
Then Area A = 2ab.
Now a² + b² = r²
so a = √(r² - b²)
So A = 2b√(r² - b²)
Now dA/db = 2√(r² - b²) + 2b *½ * -2b/√(r² - b²)
= 2((r² - b²) - 2b²)/√(r² - b²)
= 2(r² - 2b²)/√(r² - b²)
= 0 for stationary points
ie r² - 2b² = 0
so b = r/√2
whens 2b = base length (along radius) = √(2)*r
a = √(r² - b²)
= r/√(2)
= √(2)*r/2

So dimensions are √(2)*r/2 x √(2)*r (and maximum area is r²)

Answer 2

Also possible without using calculus:

A = (r * 2sinθ)(r * cosθ)
= r^2 * 2sinθcosθ
= r^2 * sin2θ

Thus, for maximal A while r is constant, make sin2θ maximum, i.e. 2θ = 90 degrees, and θ = 45 degrees,
So, dimensions are r*2sinθ = r * 2 * 1/sqrt(2) = sqrt(2) r
and (1/sqrt(2)) r

Answer 3

A = 2rsinθ*rcosθ
dA/dθ = 2r^2(-sin^2θ + cos^2θ)
A will be max when dA/dθ = 0
sin^2θ = cos^2θ
sinθ = cosθ
tanθ = 1
θ = π/4
A = 1.414r x 0.707r

Answer 4

dimensions is2 r *r
Dimensions 2r x r