The altitude (i.e., height) of a triangle is increasing at a rate of 2.5 cm/minute while the area of the triangle is increasing at a rate of 5 square cm/minute.

Question

The altitude (i.e., height) of a triangle is increasing at a rate of 2.5 cm/minute while the area of the triangle is increasing at a rate of 5 square cm/minute. At what rate is the base of the triangle changing when the altitude is 7 centimeters and the area is 96 square centimeters?

The base is changing at  ? cm/min.

Answer 1

Solution:

A = (1/2)*b*h

b = 2A/h

d/dt [ b ] = d/dt [ 2A/h ]

db/dt = 2 * d/dt [ A*h^(-1) ]

b' = 2*[ (A')*h^(-1) + A*(-1)h^(-2)*(h') ] Use Product and chain rules

b' = 2*[ (A')*h^(-1) - (A)*h^(-2)*(h') ]

We have:

A = 96

A' = 5

h = 7

h' = 2.5

b' = 2*[ (5)*(1/7) - (96)*(1/49)*(2.5) ]

b' = 2*[ -4.1837 ]

b' = -8.3673

The base is getting smaller at a rate of 8.3673 cm/minute