Problem 21 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Consider a territorial bird that harvests only in its defended territory (assumed to be circular in shape). The amount of food available can be assumed to be proportional to the area of the territory and therefore proportional to \(d^{2},\) the square of the diameter of the territory. Assume that the food gathered is proportional to the amount of food available times the time spent gathering food. Let the unit of time be one day, and suppose the amount of time spent defending the territory is proportional to the length of the territory boundary and therefore equal to \(k \times d\) for some constant, \(k\). Then \(1-k d\) is the amount of time available to gather food, and the amount, \(F\) of food gathered will be $$F=k_{2} d^{2}(1-k d)$$ Find the value of \(d\) that will maximize the amount of food gathered.

Step-by-Step Solution

Answer

The optimal diameter \( d \) is \( \frac{2}{3k} \), maximizing food gathering.

Step 1: Define the Function to Maximize

The function for the amount of food gathered, \( F \), in terms of \( d \) is given by \( F = k_{2} d^{2} (1 - k d) \). Our goal is to find the value of \( d \) that maximizes \( F \).

Step 2: Compute the Derivative

To find the maximum value, we first need the derivative of \( F \) with respect to \( d \). Using the product rule, differentiate: \[ \frac{dF}{dd} = k_{2} \cdot \frac{d}{dd} \left( d^{2}(1-kd) \right) \] Let's expand the expression: \[ k_{2} \cdot (2d(1-kd) + d^{2}(-k)) \] Simplifying: \[ \frac{dF}{dd} = k_{2} (2d - 2kd^2 - kd^2) = k_{2} (2d - 3kd^2) \] So, \( \frac{dF}{dd} = k_{2} (2d - 3kd^2) \).

Step 3: Set the Derivative to Zero

To find the critical points, set the derivative to zero: \[ k_{2} (2d - 3kd^2) = 0 \] This implies: \[ 2d - 3kd^2 = 0 \] Solve for \( d \): \[ 2d = 3kd^2 \] \[ 2 = 3kd \] \[ d = \frac{2}{3k} \] Thus, one critical point is \( d = \frac{2}{3k} \).

Step 4: Verify Maximum Using Second Derivative

Compute the second derivative for verification: \[ \frac{d^2F}{dd^2} = \frac{d}{dd} \left( k_{2} (2d - 3kd^2) \right) \] Simplifying: \[ \frac{d^2F}{dd^2} = k_{2} (2 - 6kd) \] Substituting \( d = \frac{2}{3k} \): \[ \frac{d^2F}{dd^2} = k_{2} (2 - 6k \cdot \frac{2}{3k}) = -2k_{2} \] Since \(-2k_{2}\) is negative, the critical point is a maximum.

Key Concepts

Derivatives and Critical Points

Optimizing the amount of food gathered by the territorial bird involves understanding how changes in the diameter, \( d \), of the territory affect the collected food. This scenario requires the use of derivatives, as they inform us about the rate of change of the function – in this case, the function \( F = k_{2} d^2 (1 - kd) \), representing food gathered. To find where \( F \) is maximized, we first compute its derivative with respect to \( d \), which is done using the product rule of differentiation.

The derivative \( \frac{dF}{dd} = k_{2} (2d - 3kd^2) \) provides us with critical points when it is set to zero. Finding critical points is crucial because these points can indicate potential maxima or minima where the function's value is either a peak or a trough. For our problem, we set the expression \( 2d - 3kd^2 = 0 \). Solving this equation gives us \( d = \frac{2}{3k} \), a critical point to investigate further.

The significance of finding critical points is tied closely to the optimization process in calculus. These points allow us to properly evaluate and confirm that a function is indeed maximized or minimized under given constraints, helping us make informed decisions.

Second Derivative Test

After identifying the critical point \( d = \frac{2}{3k} \) through the first derivative, the second derivative test helps ascertain whether this point is indeed a maximum. The utility of the second derivative is indicative of the curve's concavity at the critical point. When it's negative, it implies the curve is concave down, indicating a local maximum.

In our problem, computing the second derivative gives \( \frac{d^2F}{dd^2} = k_{2} (2 - 6kd) \). By substituting the critical point \( d = \frac{2}{3k} \) into this expression, we determine \( \frac{d^2F}{dd^2} = -2k_{2} \). This negative result confirms the point is indeed a maximum, reinforcing our hypothesis from the first derivative test.

This test is instrumental in ensuring that the identified critical point does not just represent any change in the function but the highest possible amount of food. Thus, the second derivative is key for verifying the maximum value problem, encapsulating a core part of optimization strategies in calculus.

Mathematical Modeling

To frame a realistic mathematical model, we need to capture all the dynamic interactions within an animal's territory. In this case, the territorial bird’s food gathering is simulated by analyzing its circular territory, where the food amount is noted as \( F = k_2 d^2 (1 - kd) \).

In mathematical modeling, taking real-world scenarios and simplifying them into solvable equations requires careful assumptions. Here, assuming that food availability depends on the territory's size and that defense time affects gathering time simplifies the complexities into a manageable function. This model allows us to apply calculus techniques, like finding derivatives, to determine an optimal strategy.

Such models serve as a bridge between theory and practice, allowing us to apply calculus to solve tangible problems. By focusing on the essential elements of the scenario into our equations, we can understand how variables interact, leading to practical solutions for maximizing outcome – in our case, the food collected by the bird.