Problem 41 - Applied Calculus

Question

The oxygen supply, \(S\), in the blood depends on the hematocrit, \(H,\) the percentage of red blood cells in the blood: \(S=a H e^{-b H} \quad\) for positive constants \(a, b\) (a) What value of \(H\) maximizes the oxygen supply? What is the maximum oxygen supply? (b) How does increasing the value of the constants \(a\) and \(b\) change the maximum value of \(S ?\)

Step-by-Step Solution

Answer

(a) H = 1/b; Maximum S = (a/b)e^{-1}. (b) Increasing 'a' increases S, while increasing 'b' decreases S.

Step 1: Understand the Function

We are given the function for oxygen supply, \(S = a H e^{-b H}\), where \(a\) and \(b\) are positive constants. Our goal is to find the value of \(H\) that maximizes this function.

Step 2: Take the Derivative

To find the maximum, we start by taking the derivative of \(S\) with respect to \(H\). Using the product rule and chain rule, we have:\[\frac{dS}{dH} = a e^{-bH} - abH e^{-bH}\] Factor out \(a e^{-bH}\):\[\frac{dS}{dH} = a e^{-bH} (1 - bH)\]

Step 3: Set Derivative to Zero

To find the critical points, set the derivative equal to zero:\[a e^{-bH} (1 - bH) = 0\]Since \(a e^{-bH} eq 0\) for any value of \(H\), we solve:\[1 - bH = 0\]

Step 4: Solve for Critical Point

From the expression \(1 - bH = 0\), solve for \(H\):\[H = \frac{1}{b}\]This is the critical point and potentially the point where \(S\) is maximized.

Step 5: Determine Maximum Supply

Substitute \(H = \frac{1}{b}\) back into the original function to find the maximum oxygen supply:\[S_{max} = a \left(\frac{1}{b}\right) e^{-b\left(\frac{1}{b}\right)} = \frac{a}{b} e^{-1}\]

Step 6: Analyze Impact of Constants

Consider how changes in \(a\) and \(b\) affect \(S_{max}\):- Increasing \(a\) linearly increases \(S_{max}\) since it scales the entire function.- Increasing \(b\) linearly decreases \(S_{max}\) as it is inversely proportional within \(\frac{a}{b}\).

Key Concepts

Derivative

In calculus, the derivative of a function represents the rate at which the function's value changes with respect to changes in its input value. For the given function of oxygen supply, \( S = aH e^{-bH} \), finding the derivative helps us understand how the oxygen supply changes with respect to hematocrit, \( H \). Derivatives are a fundamental tool for optimization because they provide the necessary information to locate maxima or minima of a function. To find the derivative of \( S \) with respect to \( H \), we identify how changes in \( H \) impact \( S \). By using derivative rules such as the product rule and the chain rule, we pinpoint where changes in the function occur most significantly.

Critical Points

Critical points in calculus are the values of \( x \) at which the derivative of a function is zero or undefined. At these points, the function may achieve local maxima, local minima, or saddle points. In the given problem, after obtaining the derivative \( \frac{dS}{dH} = a e^{-bH} (1 - bH) \), we identify critical points by setting the derivative equation to zero: \( 1 - bH = 0 \). Solving for \( H \), we find \( H = \frac{1}{b} \). This critical point helps us determine whether it's a maximum or minimum, typically through further analysis or by evaluating the second derivative. For this function, \( H = \frac{1}{b} \) is evaluated as a maximum since it yields the highest oxygen supply.

Product Rule

The product rule is a technique used in calculus to differentiate functions that are products of two or more functions. When given two functions, \( u(x) \) and \( v(x) \), the derivative of their product, \( uv \), is expressed as \( u'v + uv' \). In our oxygen supply example, the function \( S = aH e^{-bH} \) is a product of two functions: \( aH \) and \( e^{-bH} \). Applying the product rule requires that we differentiate each function separately and combine according to the rule: - The derivative of \( aH \) with respect to \( H \) is \( a \). - The derivative of \( e^{-bH} \) with respect to \( H \) involves the chain rule, resulting in \( -abe^{-bH} \). Combining these results according to the product rule gives \( \frac{dS}{dH} = a e^{-bH} - abH e^{-bH} \), which further factors to \( a e^{-bH} (1 - bH) \).

Chain Rule

The chain rule is a fundamental concept in calculus for finding the derivative of composite functions. It allows us to differentiate a function that consists of another function. If you have a composite function \( f(g(x)) \), then its derivative is \( f'(g(x)) \cdot g'(x) \). In the context of the oxygen supply function \( S = aH e^{-bH} \), the term \( e^{-bH} \) is a composite function. The outer function is the exponential and the inner function is the linear \( -bH \). To differentiate \( e^{-bH} \) with respect to \( H \), we first differentiate the outer function \( e^u \) with respect to \( u \) (where \( u = -bH \)), resulting in \( e^u \). Then, we multiply by the derivative of the inner function \( u = -bH \), which is \( -b \). Combining these gives us \( -b e^{-bH} \), allowing us to use this in combination with the product rule for the full derivative calculation.