altitude triangle help

Question

The altitude of a triangle is increasing at a rate of  centimeters/minute while the area of the triangle is increasing at a rate of  square centimeters/minute. At what rate is the base of the triangle changing when the altitude is  centimeters and the area is  square centimeters?

Answer 1

A= bh/2. 85 = 10b/2 =5b, b = 85/5 = 17

2A = bh

2A' = bh' + hb'

2(3.5) = 17(1) + 10b'

7= 17+ 10b'

10b' =-10

b' =-1 cm/min

Answer 2

This is a related rates problem.

Let A = area, h = altitude and b = base.

Since A = (1/2) b h ; b = 2A / h

By the chain rule and power rules:

db/dt = 2 (dA/dt)/ h - (2A/ h²) (dh /dt )

Plugging in dA/dt = 3.5 ; dh / dt = 1; A = 85 and h = 10 one gets

db/dt = -1 (the unit will be cm/s)

Answer 3

A = bh/2; 2dA/dt = bdh/dt + hdb/dt;

85 = 10b/2; b = 17 cm

db/dt = (2dA/dt - bdh/dt)/h = (2·3.5 cm²/min - 17cm·1cm/min)/10cm = - 1 cm/min