The altitude (i.e., height) of a triangle is increasing at a rate of 2 cm/minute while the area of the triangle is increasing at a rate of 4 square cm/minute. At what rate is the base of the triangle changing when the altitude is 10.5 centimeters and the area is 97 square centimeters?

Answer 1

Area of triangle:
A = (1/2)bh

When A = 97 and h = 10.5,
A = (1/2)bh
97 = (1/2)(b)(10.5)
b = 18.48

Differentiate area with respect to time:
A = (1/2)bh
dA/dt = (1/2)( h*db/dt + b*dh/dt )
Substitute the values:
4 = (1/2)( 10.5*db/dt + 18.48*2)

Solve for db/dt.
hope this helps~ `u`