Brine Tank Optimization Problem

[tornzaer]

Homework Statement

A tank is initially filled with 1000 litres of brine, containing 0.15 kg of salt per litre. Fresh brine containing 0.25 kg of salt per litre runs into the tank at the rate of 4 litres per second, and the mixture (kept uniform by stirring) runs out at the same rate. Show that if Q (in kg) is the amount of salt in the tank at the time t (in seconds) then d/dt Q(t) = 1 - ((Q(t))/(250))

Homework Equations

I know I have to take the derivative since this is an optimization problem and it includes rates, but nothing else.

The Attempt at a Solution

I know that the there are 1000 litres of brine and 0.15 kg of salt per litre. the 0.15 kg is already in. In addition, 0.25 kg of salt per litre runs into the tank at a rate of 4 litres per second. This is where I'm having trouble. How do I formulate an equation in relation to the one given?

Please help.

 

[Mark44]

Some of what you know is not true. You do not have to take a derivative, nor is this an optimization problem. Your job in this problem is to show that the differential equation dQ/dt = 1 - Q/250 models the situation described in your problem.

Q(t) = quantity of salt in kg at time t.
It is not true that there is .15 kg at time t = 0. You need to write an equation that expresses the amount of salt coming into the tank per minute and the amount going out per second. For example, a brine solution of .25 kg of salt comes into the tank at a rate of 4 liters/sec. This means that salt is coming into the tank at a rate of .25 kg/L*4 L/sec = 1 kg/sec.

 

[tornzaer]

Alright. How would I go about setting up the equation then? I'm still not clear on that.

 

[Mark44]

What quantity of salt is coming in per second? What quantity of salt is going out per second. The rate in - the rate out is dQ/dt. If more is coming in than going out, dQ/dt > 0. If more is going out than coming in, dQ/dt < 0.