The altitude of a triangle is increasing at a rate of 1.500 centimeters/minute while the area of the triangle is increasing at a rate of 3.000 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 9.500 centimeters and the area is 84.000 square centimeters?

Let A = the triangle's area

Let b = its base

Let h = its height (altitude)

A = (1/2)*b*h

The rate that the area is changing is dA/dt. Use the Product Rule to evaluate the derivative.

A' = (1/2)b*h' + (1/2)*b'*h

Where A'=dA/dt, h'=dh/dt, and b'=db/dt. Now solve for b', the rate that the base is changing:

A' - (1/2)*b*h' = (1/2)*h*b'

**(2A'/h) - (b*h'/h) = b'**

Now plug in the specific values from your problem statement to find the specific value of b'.

A' = 3 cm

^{2}/minh = 9.5 cm

h' = 1.5 cm/min

b = 2A/h = 2(84)/(9.5) cm