Related rates altitude problem

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 3.5 square cm/minute. At what rate is the base of the triangle changing when the altitude is 10.5 centimeters and the area is 81 square centimeters?

Answer 1

Area = bh/2 = A

A'= (1/2)(bh' +hb')

3.5 = .5(2.5b + 10.5b')

find b, plug it in above and solve for b'= rate of change of the base

A= bh/2 =b(10.5)/2

2(81)/10.5 = b

b = 162/10.5

2.5(162/10.5) + 10.5b' = 7

5(162/21) + 10.5b' = 7

270/7+10.5b'=7

10.5b' = 49/7 - 270/7 = -221/7

b'= -221/7(10.5) = -442/147=

about -3 cm/sec = rate of change of the base

when A=81 cm^2 and altitude is 10.5 cm, altitude increasing 2.5 cm/sec

and Area increasing 3.5 centimeters per second

one second later, A goes from 81 to about 84.5

altitude goes from 10.5 to about 13

base goes from 2(81)/10.5 to about 15.43 to 2(84.5/13 = 12.1 or a change of about -3 cm