1. ## related rates question

ok so i have this rate of change problem, driving me mad . its probably because of my algebra, but lets see
it is the altitude of a triangle decreases at a rate of 2cm/s while the base increases at a rate of 4 cm/s. how fast is the area changing when the altitude is 25 cm and the base is 10 cm.
what i have is dh/dt is 2cm/s and db/dt is 4
da/dt | b=10, h=25

a= 1/2 bh
a=bh/2
da/dt= ((db/dt*h + dh/dt*b)2 - (0)bh )/2
after canceling the 2's and subing in i get da/dt=(4)25+2(10)
da/dt = 120 cm^2/s
the answer says its 40 cm ^2/s not sure what im doin wrong here

2. ive never seen a question like this before, so sorry if i am wrong.

$\frac{da}{dt} = \frac {d(0.5bh)}{dt} = 0.5 \frac{d(bh)}{dt}$

$= 0.5 \left( h \frac{db}{dt} + b \frac{dh}{dt} \right)$ (product rule)

$= 0.5(25*4 + 10*-2) = 40$