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At what rate is the base of the triangle changing

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?

Philip P. | Effective and Affordable Math TutorEffective and Affordable Math Tutor
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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 cm2/min
h = 9.5 cm
h' = 1.5 cm/min
b = 2A/h = 2(84)/(9.5) cm