We are tasked with finding the absolute maximum and absolute minimum of the function:

$g(t) = t\sqrt{4 - t}$

on the closed interval $[0.9, 2.7]$.

Step 1: Identify critical points

To find the critical points, we first need to compute the derivative of $g(t)$ and set it equal to zero.

$g(t) = t(4 - t)^{1/2}$

We'll use the product rule to differentiate $g(t)$.

The product rule states: $(f \cdot h)' = f' \cdot h + f \cdot h'$

Here, $f(t) = t$ and $h(t) = (4 - t)^{1/2}$.

Derivatives:

$f'(t) = 1$ $h'(t) = \frac{d}{dt}\left( (4 - t)^{1/2} \right) = -\frac{1}{2}(4 - t)^{-1/2}$

Now applying the product rule: $g'(t) = 1 \cdot (4 - t)^{1/2} + t \cdot \left(-\frac{1}{2}(4 - t)^{-1/2}\right)$ $g'(t) = (4 - t)^{1/2} - \frac{t}{2(4 - t)^{1/2}}$

We can rewrite this expression with a common denominator: $g'(t) = \frac{2(4 - t) - t}{2(4 - t)^{1/2}} = \frac{8 - 3t}{2(4 - t)^{1/2}}$

Step 2: Solve $g'(t) = 0$

To find the critical points, set $g'(t) = 0$:

$\frac{8 - 3t}{2(4 - t)^{1/2}} = 0$

This implies: $8 - 3t = 0$ $t = \frac{8}{3} \approx 2.67$

Step 3: Evaluate $g(t)$ at critical points and endpoints

The critical point is $t = \frac{8}{3} \approx 2.67$, and the endpoints are $t = 0.9$ and $t = 2.7$. Now, evaluate $g(t)$ at these points:

• $g(0.9) = 0.9 \sqrt{4 - 0.9} = 0.9 \sqrt{3.1} \approx 0.9 \times 1.76 \approx 1.584$
• $g(2.67) = 2.67 \sqrt{4 - 2.67} = 2.67 \sqrt{1.33} \approx 2.67 \times 1.15 \approx 3.071$
• $g(2.7) = 2.7 \sqrt{4 - 2.7} = 2.7 \sqrt{1.3} \approx 2.7 \times 1.14 \approx 3.078$

Step 4: Determine absolute maximum and minimum

• $g(0.9) \approx 1.584$
• $g(2.67) \approx 3.071$
• $g(2.7) \approx 3.078$

The absolute maximum occurs at $t = 2.7$, where $g(t) \approx 3.078$.

The absolute minimum occurs at $t = 0.9$, where $g(t) \approx 1.584$.

Final Answer:

• Absolute maximum: $g(2.7) \approx 3.078$
• Absolute minimum: $g(0.9) \approx 1.584$

Would you like more details on any step or have further questions?

Here are 5 related questions to consider:

1. How do we apply the product rule in calculus?
2. Why do we need to check the endpoints in optimization problems on a closed interval?
3. Can there be more than one critical point in such problems?
4. What is the importance of common denominators when simplifying expressions?
5. How does the square root function affect the behavior of the derivative?

Tip: Always verify your critical points by plugging them back into the original function to avoid missing key points in optimization problems!