Find the maximum area of a trapezoid inscribed in a semicircle
1 answer:
Without loss of generality, we can assume the semicircle has a radius of 1 and is described by
y = √(1 - x²)
Then the shorter base has length 2x and the longer base has length 2. The area of the trapezoid is
A = (1/2)(2x+2)√(1-x²) = (1+x)√(1-x²)
Differentiating with respect to x, we have
A' = √(1-x²) + (1+x)(-2x)/(2√(1-x²)
Setting this to zero, we get
0 = (1-x²) +(1+x)(-x)
0 = 2x² +x -1
(2x-1)(x+1) = 0
x = {-1, 1/2} . . . . . -1 is an extraneous solution that gives minimum area
So, for x = 1/2, the area is
A = (1 + 1/2)√(1 - (1/2)² = (3/2)√(3/4)
A = (3/4)√3
Of course, if the radius of the semicircle is "r", the maximum area is
A = (r²·3·√3)/4
y = √(1 - x²)
Then the shorter base has length 2x and the longer base has length 2. The area of the trapezoid is
A = (1/2)(2x+2)√(1-x²) = (1+x)√(1-x²)
Differentiating with respect to x, we have
A' = √(1-x²) + (1+x)(-2x)/(2√(1-x²)
Setting this to zero, we get
0 = (1-x²) +(1+x)(-x)
0 = 2x² +x -1
(2x-1)(x+1) = 0
x = {-1, 1/2} . . . . . -1 is an extraneous solution that gives minimum area
So, for x = 1/2, the area is
A = (1 + 1/2)√(1 - (1/2)² = (3/2)√(3/4)
A = (3/4)√3
Of course, if the radius of the semicircle is "r", the maximum area is
A = (r²·3·√3)/4
You might be interested in