Quiz 3
Foundations of Calculus
Name: Adrian Reed D. Lagarde
Score: ____________
Course, Year and Section: MAEd Math
Date: _____________
1. An isosceles trapezoid is inscribed in a semi-circle of radius 20 cm so that the longer base is
along the diameter of the semi-circle. Find the length of the shorter base if the area of the
trapezoid is to be a maximum.
Solution:
An isosceles trapezoid with maximum area in this case will have one of its bases (𝑏2 ) equal to
the diameter (2𝑟) of the semicircle. Denoting 𝑥 as the 𝑏1 of the trapezoid, then the area can be
expressed into:
ℎ(𝑥 + 40)
𝐴(𝑥) =
2
Using Pythagorean theorem to derive the height of the trapezoid (refer to the illustration below):
𝑥
𝑥 2
ℎ2 = (20)2 − ( )
2
2
1600
−
𝑥
ℎ2 =
4
ℎ=√
1600 − 𝑥 2
4
Thus,
𝐴=
(𝑥+40)
2
400−𝑥 2
(√
4
1
) or 𝐴 = 4 (𝑥 + 40)√1600 − 𝑥 2
Solve for 𝐴′ .
𝐴′ =
1
1
1
{[(𝑥 + 40) (1600 − 𝑥 2 )−2 (−2𝑥)] + [(1) (√1600 − 𝑥 2 )]}
4
2
1 (𝑥 + 40)(−𝑥)
{[
] + √1600 − 𝑥 2 }
4 √1600 − 𝑥 2
1 (𝑥 + 40)(−𝑥)
𝐴′ = (
+ √1600 − 𝑥 2 )
4 √1600 − 𝑥 2
𝐴′ (𝑥) =
𝐴′ = −
𝑥 2 + 20𝑥 − 800
√1600 − 𝑥 2
For maximum area, A’ has t be 0 or 𝐴′ = 0.
−
𝑥 2 + 20𝑥 − 800
√400 − 𝑥 2
𝑥 2 + 20𝑥 − 800
√400 − 𝑥 2
=0
=0
𝑥 2 + 20𝑥 − 800 = 0
We can solve for the values of 𝑥 .
𝑥 2 + 20𝑥 − 800 = 0
(𝑥 = −20)(𝑥 + 40) = 0
𝑥 = 20 ; 𝑥 = −40
Since 𝑥 > 0, this implies that 𝑥 = 20 or the shorter base is equal to 20 𝑐𝑚 .