The altitude of a triangle is increasing at a rate of 2.5 centimeters/minute while the area of the triangle is increasing at a rate of 1 square centimeters/minute.

Question

At what rate is the base of the triangle changing when the altitude is 11.5 centimeters and the area is 95 square centimeters?

Answer 1

Let A = area, b = base, and h = height

A = (1/2)(b)(h)

Given: dA/dt = 1 and dh/dt = 2.5

Find: db/dt when h = 11.5 and A = 95

Using the product rule, dA/dt = (1/2)[b(dh/dt) + h(db/dt)]

When A = 95 and h = 11.5, 95 = (1/2)(11.5)b. So, b = 16.5217.

Therefore, 1 = (1/2)[16.5217(2.5) + 11.5(db/dt)]

db/dt ≈ -3.42 cm/min (the base is decreasing at the rate of 3.42 cm/min when A = 95 cm² and h = 11.5 cm).