Question

The altitude of a triangle is increasing at a rate of 2000 centimeters/minute while the area of the triangle is increasing at a rate of 2000 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 8000 centimeters and the area is 96000 square centimeters?

Answer 1

from the text of your problem, we can write:
\[\frac{ dh }{ dt }=2000, \frac{ dS }{ dt }=2000\]
where h is the altitude of your triangle, and S is the area of that triangle, t is time
Now, we have 2S=b*h,
where b is the base of your triangle.
Differentiating the preceeding formula for S, we get:
\[2\frac{ dS }{ dt }=\frac{ db }{ dt }*h+b*\frac{ dh }{ dt }\]
Please try to substitute your numerical data into the above equation, and you will get the rate for b. Please write youe answer

Answer 2

I got db/dt=570 cm/min