Certain chemotherapy dosages depend on a patient's surface area. According to the Mosteller model, $\mathrm{S}=\frac{\sqrt{\mathrm{hw}}}{60}$, where $\mathrm{h}$ is the patient's height in centimeters, $\mathrm{w}$ is the patient's weight in kilograms, and S is the approximation to the patient's surface area in square meters. Assume that Kim's height is a constant $161 \mathrm{~cm}$, but she is losing weight. If she loses $5 \mathrm{~kg}$ per month, how fast is her surface area decreasing at the instant she weighs $84 \mathrm{~kg}$ ?
Use implicit differentiation to determine $\frac{\mathrm{dS}}{\mathrm{dt}}$ when Kim's height is a constant $161 \mathrm{~cm}$.
\[
\frac{d S}{d t}=\square \frac{d w}{d t}
\]

Step 1 ：Given that Kim's height is a constant 161 cm, and she is losing weight at a rate of 5 kg per month, we are asked to find how fast her surface area is decreasing at the instant she weighs 84 kg.

Step 2 ：We use the Mosteller model for estimating body surface area, which is given by \(S = \frac{\sqrt{hw}}{60}\), where \(h\) is the height in cm, \(w\) is the weight in kg, and \(S\) is the surface area in square meters.

Step 3 ：Since Kim's height is constant, its derivative with respect to time is zero. We differentiate the formula for \(S\) with respect to time \(t\) to find \(\frac{dS}{dt}\).

Step 4 ：Using the chain rule, we get \(\frac{dS}{dt} = \frac{1}{2} \frac{\sqrt{h}}{\sqrt{w}} \frac{dw}{dt}\).

Step 5 ：Substituting the given values, we have \(h = 161\), \(w = 84\), and \(\frac{dw}{dt} = -5\).

Step 6 ：Substituting these values into the derivative, we find \(\frac{dS}{dt} = -\frac{\sqrt{161}}{24\sqrt{84}}\).

Step 7 ：Simplifying this expression, we find \(\frac{dS}{dt} = -\frac{\sqrt{69}}{144}\).

Step 8 ：Thus, the rate at which Kim's surface area is decreasing at the instant she weighs 84 kg is \(\boxed{-\frac{\sqrt{69}}{144}}\) square meters per month.