calculus question that I need help with

Question

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 1 cm²/min. At what rate is the base of the triangle changing when the altitude is 10 cmand the area is 300 cm²? (Round your answer to one decimal place.)

Answer 1

A=hb/2= Area of a triangle= half the base times the height

take the derivative with respect to time, A'=dA/dt, h'=dh/dt, b'=db/dt

A'=1, h'=1, A=300, h=10, b=60

height =10, Area =300 means 300=bh/2 = 10b/2, 10b=600, b=600/10=60

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

plug in the known values for AA',h,h' and b to get:

(1)=(1/2)(10b'+60(1))

solve for b'

1=5b'+30

5b'+30=1

divide by 5

b' +6 =1/5

b'=-5.8

b' = -5.8 cm/min= rate of change of the base

the Area inceased only 1 cm^2 when h increases by 1 cm

in one minute height will increase to 11 cm, and area will increase to 301 cm^2

at that time, in one minute the base will decrease to 54 8/11 from 60

so b' should be in the range of 60-54 8/11 = -5 3/11 = about - 5.3 per minute

or look at the situation 1 minute earlier, height was 9 cm and Area was 299 cm^2

then one minute earlier base = 66 4/9 base is decreasing in that 1 minute by 6 4/9 = about 6.4

average -5.3 and -6.4 to get about -5.8 cm/min