To solve this problem, we will use the concept of related rates in calculus. Let's analyze the given information step-by-step.

Given:

• The length of the rectangle, $l$, is increasing at a rate of $2 \, \text{cm/s}$.
• The width of the rectangle, $w$, is increasing at a rate of $3 \, \text{cm/s}$.
• When $l = 20 \, \text{cm}$ and $w = 10 \, \text{cm}$.

We need to determine the rate at which the area, $A$, of the rectangle is increasing at that instant.

The area $A$ of the rectangle is given by: $A = l \times w$

Taking the derivative of both sides with respect to time $t$: $\frac{dA}{dt} = \frac{d(l \times w)}{dt}$

Using the product rule: $\frac{dA}{dt} = l \frac{dw}{dt} + w \frac{dl}{dt}$

Now, substitute the given values:

• $l = 20 \, \text{cm}$
• $w = 10 \, \text{cm}$
• $\frac{dl}{dt} = 2 \, \text{cm/s}$
• $\frac{dw}{dt} = 3 \, \text{cm/s}$

$\frac{dA}{dt} = (20 \, \text{cm}) \left( 3 \, \text{cm/s} \right) + (10 \, \text{cm}) \left( 2 \, \text{cm/s} \right)$ $\frac{dA}{dt} = 60 \, \text{cm}^2/\text{s} + 20 \, \text{cm}^2/\text{s}$ $\frac{dA}{dt} = 80 \, \text{cm}^2/\text{s}$

So, the area of the rectangle is increasing at a rate of $80 \, \text{cm}^2/\text{s}$.

The correct answer is B) Tăng với tốc độ $8 \, \text{cm}^2/\text{s}$ (noting that the provided answer should read $80 \, \text{cm}^2/\text{s}$).

Do you have any questions or would you like more details?

Here are 5 additional questions you might want to ask:

1. How do you apply the product rule in calculus?
2. What are related rates in calculus?
3. Can you explain more about the concept of differentiation?
4. How would you solve a similar problem involving a circle's area changing with respect to its radius?
5. What are some practical applications of related rates in real life?

Tip: When dealing with related rates problems, always identify all the given rates and the quantity you need to find, and then use the appropriate differentiation rules.