calculus question?

Question

The altitude of a triangle is increasing at a rate of 3cm/min while the area of a triangle is increasing at a rate of 4cm^2/min.At what rate is the base of the triangle changing when the altitude is 5cm and the area is 65cm^2

Answer 1

dh/dt=3 cm/min
dA/dt=4 cm²/min
h=5 cm
A=65 cm²
A=bh/2
b=2A/h
db/dt=2*((dA/dt)h-(dh/dt)*A)/h²
db/dt=2*(4*5-3*65)/25
db/dt=-14 cm/min

Answer 2

A= 1/2* B*H,
Or,
B= 2*A/ H
Or,
d(B)/dt = 2 * d(A/H)/dt
= 2*( H d(A)/dt - A d(H)/dt)/H^2

Substituting values,

d(B)/dt = 2 *( 5* 4 - 65*3)/ 25
= 2 * (-175)/25
= -14 cm/min

This is the rate at which the base is changing

Answer 3

The Base of triangle will change by 26cm

Answer 4

A = 1/2bh

65 = 1/2b * 5
b = 26

61 = 1/2b * 2
b = 61

69 = 1/2b * 8
b = 17.25

previous minute:
rate b = (26 - 61) / 1 min
rate b = -35 cm/min (average)

next minute:
rate b = (17.25 - 26) / 1 min
rate b = -8.75 cm/min (average)