91Ó°ÊÓ

In this problem, we're given a temperature change function, which expresses how a patient's temperature change depends on the dosage of a drug given. The specific function provided is \[ T = \left(\frac{C}{2} - \frac{D}{3}\right) D^2 \] Here, \(T\) represents the temperature change, \(D\) represents the dosage, and \(C\) is a positive constant. This function is essential in determining how varying dosages affect the patient's temperature. By understanding this relationship, we can determine the dosage that maximizes the temperature change.

To find the dosage that maximizes temperature change, we need to maximize the given function. In calculus, this involves finding the critical points of the function. The critical points are where the function is highest or lowest. These points are found by taking the derivative of the function and setting it to zero. In our case, the temperature change function is \[ T = \left(\frac{C}{2} - \frac{D}{3}\right) D^2 \] We then find the derivative with respect to \(D\), set it to zero, and solve for \(D\). We perform these steps to ensure we're accurately finding the dosage that results in the maximum temperature change.

The derivative is a powerful tool in calculus used to find rates of change. In optimization problems, we use derivatives to find maximum or minimum values of functions. For our temperature change function, we need \[ \frac{dT}{dD} = \frac{d}{dD} \left( \left(\frac{C}{2} - \frac{D}{3}\right) D^2 \right) \] To calculate the derivative, we must apply the product rule, which is used when differentiating products of two functions. Once we have the derivative, setting it to zero and solving for \(D\) yields our critical points. After solving, we determine which point maximizes the temperature change by using the second derivative or examining the behavior of the original function.

The product rule is a fundamental method in differentiation, used when differentiating the product of two functions. The product rule states: \[ \frac{d}{dx} [u(x) v(x)] = u'(x)v(x) + u(x)v'(x) \] For our problem, we let \( u(D) = \left(\frac{C}{2} - \frac{D}{3}\right) \) and \( v(D) = D^2 \). Using the product rule on our temperature function, we get: \[ \frac{d}{dD} \left(\left(\frac{C}{2} - \frac{D}{3}\right)D^2\right) = \left(\frac{d}{dD} \left(\frac{C}{2} - \frac{D}{3}\right) \right) D^2 + \left(\frac{C}{2} - \frac{D}{3}\right) \frac{d}{dD} \left(D^2\right) \] After calculating these derivatives separately and substituting them back, we obtain: \[ \frac{dT}{dD} = -\frac{1}{3} D^2 + \left(\frac{C}{2} - \frac{D}{3}\right) 2D \] Simplifying this expression helps us find the critical points and determine the maximum temperature change.