1. ## Question on differentiation/integration...

Here is the question...

A 100 litre tank is initially full of a mixture of 10% alcohol and 90% water. Simultaneously, a pump drains the tank at 4 litres/second, while a mixture of 80% alcohol and 20% water is poured in at rate 3 litres/second. Thus the tank will be empty after 100 seconds. Assume that the two liquids mix thoroughly, and let y litres be the amount of alcohol in the tank after t seconds.

dy/dt = 2.4 - ( 4y / (100 - t) )

Find y as a function of t. Hence deduce that the maximum amount of alcohol in the tank occurs after about 34 seconds, and is about 39.5 litres.

* I have identified the integrating factor as A = 4 / (100 - t)

Thank you so so much if you can help!!

2. Originally Posted by reneewilliams
Here is the question...

A 100 litre tank is initially full of a mixture of 10% alcohol and 90% water.
So y(0)= .1(100)= 10 litres.

Simultaneously, a pump drains the tank at 4 litres/second, while a mixture of 80% alcohol and 20% water is poured in at rate 3 litres/second. Thus the tank will be empty after 100 seconds. Assume that the two liquids mix thoroughly, and let y litres be the amount of alcohol in the tank after t seconds.

dy/dt = 2.4 - ( 4y / (100 - t) )

Find y as a function of t. Hence deduce that the maximum amount of alcohol in the tank occurs after about 34 seconds, and is about 39.5 litres.

* I have identified the integrating factor as A = 4 / (100 - t)

Thank you so so much if you can help!!
Good! Since you have an integrating factor, you can solve for y(t). (Don't forget to use y(0)= 10 to determine the constant of integration.) Once you have done that, differentiate y with respect to t and set equal to 0 just as you would with any function to find where there is a maximum or minimum.

3. I've used the integrating factor but am still really struggling! I have...

y / (100 - t)^4 = (integration sign) 2.4 / (100 - t)^4

What should I end up with now??

Thank you for your help so far!