(1 point) A rectangle has one side of $8 \mathrm{~cm}$. How fast is the area of the rectangle changing at the instant when the other side is $12 \mathrm{~cm}$ and increasing at 4 cm per minute? (Give units.) Answer:

Step 1 ：The area of a rectangle is given by the formula \(A = l \times w\), where \(l\) and \(w\) are the length and width of the rectangle respectively. In this case, we know that one side (let's call it \(l\)) is 8 cm and the other side (let's call it \(w\)) is 12 cm and increasing at a rate of 4 cm per minute. We want to find how fast the area is changing at this instant, which is the derivative of the area with respect to time, or \(\frac{dA}{dt}\).

Step 2 ：We can find this by using the chain rule of differentiation, which states that the derivative of a function of a function is the derivative of the outer function times the derivative of the inner function. In this case, our outer function is the area \(A = l \times w\) and our inner function is the width \(w\) with respect to time \(t\).

Step 3 ：So, we have \(\frac{dA}{dt} = l \times \frac{dw}{dt}\).

Step 4 ：We know that \(l = 8\) cm and \(\frac{dw}{dt} = 4\) cm per minute, so we can substitute these values into the equation to find \(\frac{dA}{dt}\).

Step 5 ：\(l = 8\)

Step 6 ：\(dw/dt = 4\)

Step 7 ：\(dA/dt = 32\)

Step 8 ：This means that at the instant when the width of the rectangle is 12 cm and increasing at a rate of 4 cm per minute, the area of the rectangle is increasing at a rate of 32 square cm per minute.

Step 9 ：Final Answer: The rate of change of the area of the rectangle at the given instant is \(\boxed{32 \text{ cm}^2/\text{min}}\).