the two bases of an isosceles trapezoid are 12 and 20 feet long, respectively. find the rate at which the area is changing when the equal sides are 5 feet long and are increasing at the rate of 2 feet per minute.

Answer 1

Make a sketch
I dropped perpendiculars at the ends of the 12 ft side to the 20 ft side and let the height be h ft
It is easy to see that
Area = 16h
also we know x^2 = 4^2 + h^2
h = √(x^2 - 16)
when x = 5, h = 3, and dx/dt = 2

Area = 16h = 16(x^2 - 16)^(1/2)
d(Area)/dt = 8(x^2 - 16)^(-1/2) (2x) dx/dt
which for the given data
= 8(25-16)^(-1/2)(10)(2)
= 8/√9 (20)
= 160/3

At that moment, the area is increasing at 160/3 ft^2 / min

check my arithmetic

Answer 2

Where did 16 come from?