Related Rates Problem?

Question

The area of a triangle is increasing at a rate of 4cm^2/min and its base is increasing at a rate of 1 cm/min. At what rate is the altitude of the triangle changing when the altitude is 20 cm and the area is 80 cm^2?

So far I've tried using a "BUT" statement but the answer came out to be -8 which is wrong. Also I tried looking at b throught h terms but it didn't work out. Best answer/first right answer with steps gets 10 points :)

Answer 1

A = b * h / 2
dA/dT = [b * dh/AT + h * db/dT] / 2

When h = 20 and A = 80, b = 8

4 = [8 * dh/dT + 20 * 1] / 2
4 = 4 * dh/dT + 10
-6 = 4 * dh/dT
-3/2 = dh/dT

As you can see, Gopal has the 4 * dh/dT part correct, but where Gopal has 8 minus 20, it should be 4 minus 10. As a result, Gopal's answer is twice the correct one. Gopal has a lot of good ideas, but is careless with their implementation. You can see that Gopal has unbalanced parentheses and brackets, no doubt the source of the confusion.

As a rough check, we can ask what the status is one minute after the time given in the problem, when the area is 84 and the base is 9.

h = 2 * 84 / 9

The altitude computes to 18-2/3, much closer to my rate of change than to Gopal's.

Answer 2

A=(1/2)bh
dA/dt=(1/2(hdb/dt+bdh/dt)
4=(1/2)[(1*20)+(4*dh/dt)
8-20=4dh/dt
dh/dt=-3
h decreasing at -3 cm/sec