2. (IO mards) Foraging bumblebees Many animals forage on resources that are discributed in discrete patches, for example, bumblebees visit many flowers, foraging nectar from each. The amount of nectar \( N(t) \) harvested from a given flower increases the more time is spent at the flower, but the returns diminish, he, more nectar is harvested in the first second at the flower when the nectar is most plentiful, than in the second when there is less nectar left. Suppose that these diminishing returns are described by the function, \[ N(t)=\frac{0.3 t}{t+2} \] Where \( t \) is measured in seconds and \( N \) in miligrams. Suppose also that it takes a bee 4 seconds on average to travel from one flower to the next. If a bee spends \( t \) seconds per flower then the average amount of nectar it will harvest per second from its arrival at one flower until its arrival at the next is the total nectar it harvests \( N(t) \), divided by the time taken, which is the harvest time, \( t \), plus the travel time of 4 seconds, a) Write an equation for the average amount of nectar consumed per second, \( f(t) \), b) What harvesting time per flower would maximize the average rate of energy gain, l.e. maximize the nectar gathered?

**Problem a)** The average amount of nectar harvested per second is given by the total nectar from a flower divided by the total time (harvesting time plus the 4 seconds travel time). Since the amount of nectar harvested from a flower is \[ N(t)=\frac{0.3t}{t+2}, \] if a bee spends \( t \) seconds harvesting at a flower, the average nectar gain per second is \[ f(t)=\frac{N(t)}{t+4}=\frac{\frac{0.3t}{t+2}}{t+4}=\frac{0.3t}{(t+2)(t+4)}. \] **Problem b)** To maximize \( f(t) \), we need to find the value of \( t \) that maximizes the function \[ f(t)=\frac{0.3t}{(t+2)(t+4)}. \] Since \( 0.3 \) is a constant, we can instead maximize \[ g(t)=\frac{t}{(t+2)(t+4)}. \] First, rewrite the denominator: \[ (t+2)(t+4)=t^2+6t+8. \] Then \[ g(t)=\frac{t}{t^2+6t+8}. \] Differentiate \( g(t) \) with respect to \( t \) using the quotient rule: \[ g'(t)=\frac{(1)(t^2+6t+8)-t(2t+6)}{(t^2+6t+8)^2}. \] Simplify the numerator: \[ \begin{aligned} & (t^2+6t+8)-t(2t+6) \\ & = t^2+6t+8 - (2t^2+6t) \\ & = -t^2+8. \end{aligned} \] Thus, \[ g'(t)=\frac{-t^2+8}{(t^2+6t+8)^2}. \] Set the derivative equal to zero to find the critical point: \[ -t^2+8=0 \quad \Longrightarrow \quad t^2=8 \quad \Longrightarrow \quad t=2\sqrt{2}. \] Since the denominator is always positive for \( t>-2 \) (and we are considering \( t>0 \)), the sign of \( g'(t) \) is determined by \(-t^2+8\), which changes from positive to negative as \( t \) passes \( 2\sqrt{2} \). Therefore, this critical point corresponds to a maximum. **Final Answers:** a) The average amount of nectar consumed per second is \[ f(t)=\frac{0.3t}{(t+2)(t+4)}. \] b) The harvesting time per flower that maximizes the average nectar gain is \[ t=2\sqrt{2}\ \text{seconds}. \]