Problem 32 - Applied Calculus

Question

The energy expended by a bird per day, \(E\), depends on the time spent foraging for food per day, \(F\) hours. Foraging for a shorter time requires better territory, which then requires more energy for its defense. \({ }^{4}\) Find the foraging time that minimizes energy expenditure if $$ E=0.25 F+\frac{1.7}{F^{2}}. $$

Step-by-Step Solution

Answer

The optimal foraging time is approximately 2.39 hours.

Step 1: Understand the Function

We are given the function for energy expenditure, \( E = 0.25F + \frac{1.7}{F^2} \), where \( F \) is the foraging time in hours. Our task is to find the value of \( F \) that minimizes \( E \).

Step 2: Find the Derivative

To find the minimum point, we need to compute the derivative of \( E \) with respect to \( F \). Differentiate: \[ E'(F) = \frac{d}{dF} \left(0.25F + \frac{1.7}{F^2}\right). \]This gives us:\[ E'(F) = 0.25 - \frac{3.4}{F^3}. \]

Step 3: Set Derivative to Zero

To find the critical points that might give a minimum, set the derivative \( E'(F) = 0 \): \[ 0.25 - \frac{3.4}{F^3} = 0. \] Solve for \( F \): \[ \frac{3.4}{F^3} = 0.25, \]\[ 3.4 = 0.25F^3, \]\[ F^3 = \frac{3.4}{0.25}, \]\[ F^3 = 13.6. \]

Step 4: Solve for Foraging Time

Calculate the cube root of 13.6 to find the optimal \( F \):\[ F = \sqrt[3]{13.6}. \]Use a calculator to find \( F \approx 2.39 \) hours.

Step 5: Confirm Minima

Verify that \( F = 2.39 \) is indeed a minimum by checking the second derivative \( E''(F) \):Calculate the second derivative:\[ E''(F) = \frac{d}{dF} \left(0.25 - \frac{3.4}{F^3}\right), \]\[ E''(F) = \frac{10.2}{F^4}. \]Since \( E''(F) > 0 \) for \( F > 0 \), it confirms a local minimum.

Key Concepts

Energy Expenditure

Energy expenditure is a critical concept in understanding how much energy an organism, like our bird in this example, uses in its daily activities. The function provided in this problem shows how the energy expenditure, denoted as \( E \), is affected by the time spent foraging for food, \( F \). This function models the trade-off between two components: \

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• Direct energy spent foraging (proportional to \( F \)).
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• Energy required for territory defense (inversely related to \( F^2 \)).
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Finding the optimal foraging time is key to minimizing overall daily energy expenditure, which has implications for the bird's survival and efficiency. In this context, minimizing \( E \) means balancing time spent actively searching for resources against the energy cost of defending territory when foraging time is short. This is useful both practically and fundamentally in the study of biological efficiency and ecosystem interactions.

Derivative

A derivative in calculus represents the rate at which a function is changing at any given point. In our context, the function \( E = 0.25F + \frac{1.7}{F^2} \) allows us to identify how energy expenditure changes with varying foraging time. \
To find this rate of change, we need the derivative \( E'(F) \), which tells us how much \( E \) increases or decreases as \( F \) changes. \
This derivative, \( E'(F) = 0.25 - \frac{3.4}{F^3} \), captures both aspects of the energy relationship: \

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• The consistent increase from the energy spent foraging \( (0.25F) \).
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• The rapid decrease from energy used in territory defense \( (\frac{-3.4}{F^3}) \).
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The zeroes of this derivative will identify the critical points where the energy expenditure does not change, highlighting potential optima or extrema of the function.

Critical Points

Critical points are values of \( F \) where the derivative \( E'(F) \) equals zero. These points indicate where there could be a maximum or minimum in the function \( E \). \
Finding these critical points is essential because they show where the rate of change of energy expenditure is neither increasing nor decreasing; essentially, these are the points of transition. \
To find a critical point, we solve \( 0.25 - \frac{3.4}{F^3} = 0 \). \

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• Rearranging gives \( \frac{3.4}{F^3} = 0.25 \).
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• Solving results in \( F^3 = 13.6 \), leading to \( F \approx 2.39 \) hours.
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This particular solution indicates the optimal foraging time that might minimize the energy expenditure. However, to confirm whether this critical point is a minimum or maximum, further testing is needed using the second derivative test.

Second Derivative Test

The second derivative test helps confirm whether a critical point found in a function is a local minimum or maximum. This is done by looking at the concavity of the function at that point. \
In our case, the second derivative of the energy function is \( E''(F) = \frac{10.2}{F^4} \). \

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• If \( E''(F) > 0 \), the function is concave up at \( F \), indicating a local minimum.
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• If \( E''(F) < 0 \), the function is concave down, indicating a local maximum.
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For \( F > 0 \), it's clear that \( E''(F) = \frac{10.2}{F^4} > 0 \), confirming that the critical point at \( F \approx 2.39 \) hours is indeed a local minimum. This second derivative test is a vital tool in optimization problems, providing mathematical confirmation of the nature of critical points.