Let's first compute the derivative of the given function $w(t)$ using the quotient rule.

The weight function is given by:

$w(t) = \frac{160t + 1200}{t + 6}$

Step 1: Use the Quotient Rule

The quotient rule states:

$\left( \frac{f(t)}{g(t)} \right)' = \frac{f'(t)g(t) - f(t)g'(t)}{(g(t))^2}$

Where:

• $f(t) = 160t + 1200$
• $g(t) = t + 6$

Now, we need to calculate $f'(t)$ and $g'(t)$:

$f'(t) = 160 \quad \text{(since the derivative of a linear function is its slope)}$ g'(t) = 1 \quad \text{(the derivative of t + 6 is 1)}

Step 2: Apply the Quotient Rule

Now, applying the quotient rule:

$w'(t) = \frac{(160)(t+6) - (160t+1200)(1)}{(t+6)^2}$

Simplifying the numerator:

$w'(t) = \frac{160(t+6) - (160t + 1200)}{(t+6)^2}$ $w'(t) = \frac{160t + 960 - 160t - 1200}{(t+6)^2}$ $w'(t) = \frac{-240}{(t+6)^2}$

Thus, the derivative of the weight function is:

$w'(t) = \frac{-240}{(t+6)^2}$

Step 3: Calculate $w'(12)$

Now, let's determine the value of $w'(12)$. Substitute $t = 12$ into $w'(t)$:

$w'(12) = \frac{-240}{(12+6)^2} = \frac{-240}{18^2} = \frac{-240}{324}$ $w'(12) = -\frac{20}{27}$

Interpretation in the context of the problem:

The derivative $w'(12) = -\frac{20}{27} \approx -0.74$ represents the rate of change of the person's weight after 12 weeks on the diet. Specifically, it means that at week 12, the person's weight is decreasing at a rate of approximately 0.74 pounds per week.

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Would you like further details on this calculation, or any clarification on the context?

Related Questions:

1. What is the significance of $w'(t)$ when $t = 0$?
2. How would the rate of weight loss change if the diet lasts for 40 weeks instead of 30?
3. What is the long-term behavior of $w'(t)$ as $t \to \infty$?
4. How much weight does the person lose in the first 12 weeks?
5. Can the model predict a week where the rate of weight loss is less than 0.5 pounds per week?

Tip:

Always check the unit or context when interpreting derivatives; it often provides meaningful insights into real-world changes.