A 5 ft long man walks away from foot of a 12(½) ft high lamp post at rate of 3 mph. What will be rate at which shadow increases?

Let, AB be the lamp-post whose foot is A, and B is the source of light, and given (AB)? = 12(½) ft.
Let MN denote the position of the man at time t where (MN)’ = 5ft.
Join BN and produce it to meet AM(produced) at P.
Then the length of man’s shadow= (MP)’
Assume, (AM)? = x and (MP)? = y. Then,
(PA)? = (AM)? + (MP)? = x + y
And dx/dt = velocity of the man = 3
Clearly, triangles APB and MPN are similar.
Thus, (PM)?/(MN)? = (PA)?/(AB)?
Or, y/5 = (x + y)/12(½)
Or, (25/2)y = 5x + 5y
Or, 3y = 2x
Or, y = (2/3)x
Thus, dy/dt = (2/3)(dx/dt)
As, dx/dt = 2,
= 2/3*3 = 2mph