Problem 17 - Applied Calculus

Question

For some positive constant \(C\), a patient's temperature change, \(T\), due to a dose, \(D\), of a drug is given by $$T=\left(\frac{C}{2}-\frac{D}{3}\right) D^{2}$$ (a) What dosage maximizes the temperature change? (b) The sensitivity of the body to the drug is defined as \(d T / d D\). What dosage maximizes sensitivity?

Step-by-Step Solution

Answer

(a) Dosage maximizing temperature change is \( D = C \). (b) Dosage maximizing sensitivity is also \( D = C \).

Step 1: Identify the Function

The function for temperature change due to dosage is given by \( T(D) = \left(\frac{C}{2} - \frac{D}{3}\right) D^2 \). Our goal is to find the value of \( D \) that maximizes this function.

Step 2: Find the Derivative

To find the dosage that maximizes the temperature change, we need the derivative of \( T(D) \) with respect to \( D \). This is given by using the product and power rules. Let \( T(D) = \left(\frac{C}{2}\right) D^2 - \frac{1}{3} D^3 \). The derivative is:\[ \frac{dT}{dD} = C D - D^2 \]

Step 3: Set Derivative to Zero

To find critical points, set \( \frac{dT}{dD} = 0 \):\[ CD - D^2 = 0 \]\[ D(C - D) = 0 \]This gives \( D = 0 \) or \( D = C \). We need to choose the positive, non-zero value \( D = C \) since \( D = 0 \) is not practical for maximizing temperature change.

Step 4: Second Derivative Test

Find the second derivative to confirm whether \( D = C \) is a maximum:\[ \frac{d^2T}{dD^2} = C - 2D \]Substitute \( D = C \):\[ \frac{d^2T}{dD^2}|_{D=C} = C - 2C = -C \]Since \( C > 0 \), \( -C < 0 \) indicates that \( D = C \) is a maximum.

Step 5: Analyze Sensitivity

Sensitivity \( \frac{dT}{dD} \) was already found as \( CD - D^2 \). To maximize sensitivity, find critical points of \( \frac{dT}{dD} \):The critical points occur when \( CD - D^2 = 0 \), which simplifies to \( D(C - D) = 0 \). This yields \( D = 0 \) or \( D = C \). We have concluded previously that \( D = C \) results in maximized sensitivity as this is the only non-zero practical solution.

Key Concepts

Optimization

Optimization is all about finding the best possible solution given certain constraints.
In the context of our problem, we aim to determine the drug dosage that results in the greatest change in temperature.
To do this, we use the function:

• \( T(D) = \left(\frac{C}{2} - \frac{D}{3}\right) D^2 \)

This function symbolizes how the temperature changes depending on the dosage. By finding the peak value of this function, we can pinpoint the optimal dosage.
The process involves calculating the derivative, which helps us find the function's slope.
Once the slope equals zero, it indicates that we may have found a crucial point, sometimes called a critical point.
After locating these critical values, we further assess them to determine if they correspond to a maximum (or minimum). This entire procedure is a common way to tackle optimization tasks in calculus.

Derivatives

Derivatives serve as a fundamental tool in calculus, representing the rate of change of a function.
In simpler terms, they tell us how fast or slow something is changing. In this exercise, we need to find the derivative of the given function to analyze temperature change.
Using the function:

• \( T(D) = \left(\frac{C}{2}\right) D^2 - \frac{1}{3} D^3 \)

we calculate the derivative \( \frac{dT}{dD} = C D - D^2 \).
This expression provides the rate of change of temperature with respect to dosage.
Some simple rules, like the product and power rule, guide us through finding this derivative.
The product rule is used when two functions are multiplied together, and the power rule is applied when we deal with polynomials. These rules allow us to methodically dissect complex expressions to gain meaningful insights into how changes in dosage alter temperature change.

Critical Points

Critical points play a major role in solving optimization problems. They occur where the derivative equals zero, indicating potential maximum or minimum values of a function.
For our scenario, we first solve \(CD - D^2 = 0\) to find the critical points.After factoring, we find

• \( D(C - D) = 0 \)

resulting in solutions for \( D = 0 \) or \( D = C \).
While both of these are mathematically valid, in practice, we favor the solution where \( D = C \) to maximize temperature change, as a dosage of zero wouldn't yield any useful results.
To truly confirm this as a maximum, we check the second derivative \(\frac{d^2T}{dD^2} = C - 2D\).
Substituting \( D = C \), we receive \( -C \), which is negative, confirming that this is indeed a maximum point.
Critical points act as checkpoints, giving us potential candidates for maximum or minimum values in our quests to solve real-world problems like this.