Optimization problem -- Calculus

Question

I need help with the following homework problem and setting up the equations as well as how to solve them!


(Bees collecting pollen and nectar - Optimization)
Many bees collect both pollen and nectar. Pollen is used for protein, and nectar is
used for energy.

 

Suppose the amount of nectar harvested during t seconds on a
flower is: F(t)=t/(1+t)

and that the amount of pollen harvested during t seconds on a flower is:
G(t)=t/(2+t)

.
The bee collects pollen and nectar simultaneously. Travel time between flowers is
τ = 1.0 seconds.

a) What is the optimal time for the bee to leave one flower for the next in order to
collect nectar at the maximum rate?

b) What is the optimal time for the bee to leave one flower for the next in order to
collect pollen at the maximum rate? Why are the two times different?

c) Suppose that the bee values pollen twice as much as nectar. Find a single
function V(t) that give the value of resources collected by time t. What is the
optimal time for the bee to leave? [You may need to use a computer or
calculator to solve this equation.]

d) Suppose that the bee first collects nectar and then switches to pollen. Assume it
spends 1.0 second collecting nectar. How long should it spend on pollen?

e) Suppose that the flower patch is not as populated and it takes τ = 2.5 seconds to
travel from one flower to the next. How does this impact your answers in (b),
(c), and (d)?

Answer 1

a) Suppose the bee spends x seconds on each flower. Then every x+1 seconds (to account for the travel time), it collects F(x) units of nectar, and thus averages F(x)/(x+1) units per second. To optimize this, we differentiate this with respect to x and set it equal to 0, getting (x+1)^2-2x(x+1)/(x+1)^4 = 0 with the quotient rule. This gives one solution (ignoring nonsensical negative solutions) of x=1 second per flower. (You should verify for yourself that this is a maximum, and not a minimum or a saddle point. Compute the second derivative or reason about the shape of F'(x).)

b) is the same with G instead of F. I leave that to you.

c) We can define our value function as V(t)=2G(t)+F(t), with pollen (G) valued twice as much as nectar (F). Given this, optimizing V is the same as in parts a) and b); set a variable x for the amount of time per flower, find an expression for the average rate by dividing V by x+1, and then maximize that expression by setting its derivative equal to 0.

d) is easy because one of the variables is already fixed; we are told that the bee spends one second on nectar per flower, gaining F(1)=1/2 of a unit of value. So if the bee spends x seconds getting pollen on each flower, it gains 2G(x)+1/2 value. So for each flower, it spends 1 second on nectar, 1 second traveling, and x seconds on pollen, so its rate-of-value is [2G(x)+1/2]/(x+2). Differentiate this and set it equal to 0 as before.

e) means redoing b), c), and d), but dividing G and V by x+2.5 or x+3.5 instead of x+1 or x+2, respectively.