Question

The altitude of a triangle is increasing at a rate of 1.5 centimeters/minute while the area of the triangle is increasing at a rate of 1.0 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 9.5 centimeters and the area is 97 square centimeters?

Answer 1

A = 1/2 b * h

Using the product rule we get:

dA/dt = 1/2 dh/dt * b + 1/2 db/dt * h

We know the following
dA/dt = 5
dh/dt = 1.5
db/dt = x (the thing we want to solve)
h = 11.5
b = ? (we need to find this first)

We know that A = (1/2)*b*h
97 = (1/2)*b*(11.5) we can now solve for b
97*2 = b * (23/2)
b = (97*2*2) / 23
b = 388 / 23

Now we can solve for db/dt.

dA/dt = 1/2 dh/dt * b + 1/2 db/dt * h
5 = (1/2)*(1.5)*(388/23) + (1/2)*(x)*(11.5)
5 = 291 / 23 + 23x / 4
5 - 291/23 = 23x / 4
-176 / 23 = 23x / 4
-176/23 * 4/23 = x
x = -704 / 529
x = -1.33 cm/min