1. ## Optimization!

When you cough, your windpipe contracts. The speed, v, with which air comes out depends on the radius, r, of your windpipe. If R is the normal (rest) radius of your windpipe, then for r ≤ R, the speed is given by the following formula, where a is a positive constant.

v = a(R - r)r^2

(a) Find v '(r).

v '(r) =

(b) What value of r maximizes the speed?

r =

so i believe to find the value of that r that maximizes speed, you would have to set the derivative equal to 0 and solve. However im having trouble finding the derivative, any help? thanks...

2. Originally Posted by mathlete
When you cough, your windpipe contracts. The speed, v, with which air comes out depends on the radius, r, of your windpipe. If R is the normal (rest) radius of your windpipe, then for r ≤ R, the speed is given by the following formula, where a is a positive constant.

v = a(R - r)r2

(a) Find v '(r).

v '(r) =

(b) What value of r maximizes the speed?

r =

so i believe to find the value of that r that maximizes speed, you would have to set the derivative equal to 0 and solve. However im having trouble finding the derivative, any help? thanks...
Expand the RHS of the equation:

$v(r) = aRr^2-ar^3$. ........... Calculate the first derivative:

$v'(r) = 2aRr - 3ar^2$........... and set v'(r) = 0. Solve for r.

I've got: r = 0 (this case is called strangulation) or $r = \frac23R$