A man 2 m high, walks at a uniform speed of 6 m/min away from a lamp post, 5 m high. Find the rate at which the length of his shadow increases .

Let us draw a figure as per given question.
Let AB = 5 meter and CD = 2 meter be the heights of a lamp post and the men respectively.
At any time t BD = x meter and shadow of man ED = y meter.
Then, dx/dt = 6 m/min
Now, right triangles ABE and CDE are similar, then
AB/CD = BE/DE
⇒ 5/2 = (x + y)/y
⇒ 5y = 2x + 2y
⇒ 5y - 2y = 2x
⇒ 3y = 2x
⇒ 3 dy/dt = 2 dx/dt
⇒ 3 dy/dt = 2 x 6
⇒ dy/dt = 2 x 6 / 3
∴ dy/dt = 2 x 2 = 4
Hence, length of his shadow increase at the rate of 4 m/min.