The radius of a cone increases at the rate of 0.3 inches per minute, but the volume remains constant. At what rate does the height of the cone change when the radius Is 4 inches and the height is 15 inches?

To find the rate at which the height of the cone changes, we can use related rates.

Let's denote the radius of the cone as r and the height as h.

We are given that the radius increases at a rate of 0.3 inches per minute, so dr/dt = 0.3 inches per minute.

The volume of a cone is given by V = (1/3)π(r^2)(h), and we are told that the volume remains constant. Therefore, dV/dt = 0.

We need to find dh/dt, the rate at which the height changes when r = 4 inches and h = 15 inches.

To find dh/dt, we need to find the relationship between h, r, and V using differentiation.

Taking the derivative of the volume equation with respect to time (t), we get:

dV/dt = (1/3)π[(2r)(dr/dt)(h) + (r^2)(dh/dt)]

Since dV/dt = 0, we can rewrite the equation as:

0 = (1/3)π[(2r)(0.3)h + (r^2)(dh/dt)]

Simplifying the equation:

0 = (2/3)πrh(0.3) + (1/3)π(r^2)(dh/dt)

We can substitute the given values r = 4 and h = 15 into the equation and solve for dh/dt:

0 = (2/3)π(4)(15)(0.3) + (1/3)π(4^2)(dh/dt)

0 = 120π(0.3) + 16π(dh/dt)

0 = 36π + 16π(dh/dt)

Simplifying further:

16π(dh/dt) = -36π

dh/dt = -36π / (16π)

dh/dt = -9/4

Therefore, the rate at which the height of the cone changes when r = 4 inches and h = 15 inches is -9/4 inches per minute.