Finding the maximum size of an Initial Value Problem coefficient

[MathMan2022]

Homework Statement

find the maximum size of a IVP coefficient size?

Relevant Equations

diff(T(x), x) = v/200*(45 - T(x)) + 0.015*(22 - T(x)) where T(0)=39

The following IVP
diff(T(x), x) = v/200*(45 - T(x)) + 0.015*(22 - T(x)) where T(0)=39

Describes the tempetatur T in celsius at the time x of a tub filled with water. A tub which is filled with hot water at rate of v l/min.

Lets say I am told that a guy takes a 40 min bath, and during those 40 min he doesn't want the water in the tub to become colder than 37 c.
How would I go about calculating the maximum size of v for that periode?

 

[berkeman]

Homework Statement:: find the maximum size of a IVP coefficient size?
Relevant Equations:: diff(T(x), x) = v/200*(45 - T(x)) + 0.015*(22 - T(x)) where T(0)=39

The following IVP
diff(T(x), x) = v/200*(45 - T(x)) + 0.015*(22 - T(x)) where T(0)=39

Describes the tempetatur T in celsius at the time x of a tub filled with water. A tub which is filled with hot water at rate of v l/min.

Lets say I am told that a guy takes a 40 min bath, and during those 40 min he doesn't want the water in the tub to become colder than 37 c.
How would I go about calculating the maximum size of v for that periode?

Welcome to PF.

Can you show us your work so far? We are not allowed to help you until we see your initial work. Thanks.

 

[MathMan2022]

Welcome to PF.

Can you show us your work so far? We are not allowed to help you until we see your initial work. Thanks.

Sure, If I do a dsolve in Maple on the IVP I get the following solution seen in the screenshot. But how should I got about finding the maximum size of v? If T(40)=37? Thats what I am unsure on.