The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2 cm^2/min.

Question

At what rate is the base of the triangle changing when the altitude is 20 cm and the area is 180 cm^2?

Answer 1

Area= base×height÷2 or A=bh/2

When h=20, A=20, b=2A/h=2(180)/20=18 and dh/dt = 1

dA/dt = (h db/dt + b dh/dt)/2

2 = (20 db/dt + 18(1))/2

20 db/dt = 4-18

db/dt = -14/20 = -7/10 cm/min

Answer 2

Let x = base at time t and y = altitude at time t and A = area at time t.

Given: dy/dt = 1 and dA/dt = 2

Find: dx/dt when y = 20 and A = 180

A = (1/2)xy When y = 20 and A = 180, 180 = (1/2)(x)(20), so x = 18.

So, dA/dt = (1/2)[x(dy/dt) + y(dx/dt)]

2 = (1/2)[(18)(1) + (20)(dx/dt)]

2 = 9 + 10(dx/dt)

So, dx/dt = -0.7 cm/min

The base is decreasing at the rate of 0.7 cm/min at the time in question.