Question

Milk is being pumped into a 10-foot-tall cylindrical container at a constant rate.The depth of the milk is increasing linearly.• At 1:30 p.m., the milk depth was 2.4 feet.• It is now 4:00 p.m., and the depth of the milk is 3.9 feet.What will the depth (in feet) of the milk be at 5:00 p.m.?

Answer 1

Let:

t1 = Initial time = 0

t2 = Final time = 4:00 - 1:30 = 2:30 = 2.5h

h1 = initial height = 2.4ft

h2 = final height = 3.9ft

so:

`m=(h2-h1)/(t2-t1)=(3.9-2.4)/(2.5-0)=(1.5)/(2.5)=0.6`

The rate is 0.6ft/h

Therefore, the depth of the milk at 5:00 pm is:

`3.9ft+0.6ft=4.5ft`