rate of change?

Question

The altitude of a triangle is increasing at a rate of 1000 centimeters/minute while the area of the triangle is increasing at a rate of 1000 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 8500 centimeters and the area is 83000 square centimeters?

Answer 1

Is this a calculus problem? if so:

let h = altitude, A = area.

dh/dt = 1000 [ cm/min ], dA/dt = 1000 [ cm^2/min ]

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

dA/dt = (1/2)(db/dt)h + (1/2)(dh/dt)(b)

dA/dt = (1/2)(db/dt)h + (1/2)(dh/dt)(2A/h)

1000 = (1/2)(db/dt)(8500) + (1/2)(1000)(2*83000/8500)

Now you need to solve for db/dt :)

Answer 2

Let h be the altitude and b be the base.

Then A =(1/2)bh
and dA/dt = (1/2)b dh/dt + (1/2)h db/dt

Plug in dA/dt = 1000, dh/dt = 1000, h=8500 and A=83000. Also plug in b = 2A/h = 146000/8500. You get

1000 = 146000/17000*1000 + 8500/2 db/dt

and solve for db/dt.

Answer 3

A = (1/2)bh
dA/dt = (1/2)(bdh/dt + hdb/dt)
db/dt = (2dA/dt - bdh/dt)/b
db/dt = h(2dA/dt - (2A/h)dh/dt) / 2A
db/dt = (hdA/dt - Adh/dt) / A
db/dt = (h/A)dA/dt - dh/dt
db/dt = (8500/83000)1000 - 1000
db/dt = - 1000(1 - 85/830)
db/dt ≈ - 897.590 cm/min