The length of the rectangle is changing at a rate of 4 cm/s and the area is changing at the rate of 8 cm/s. What will be the rate of change of width if the length is 4cm and the width is 1 cm

The correct option is 1 cm/s.

To find the rate of change of the width of the rectangle, we can use the formula for the area of a rectangle:
$$A=l\times w$$
where $A$ represents the area, $l$ represents the length, and $w$ represents the width of the rectangle.

Since the dimensions are changing with respect to time ($t$), we differentiate both sides of the equation with respect to $t$ using the product rule:
$$\frac{dA}{dt}=l\cdot \frac{dw}{dt}+w\cdot \frac{dl}{dt}$$

From the problem description, we are given the following values:
- The rate of change of length, $\frac{dl}{dt}=4\text{cm/s}$
- The rate of change of area, $\frac{dA}{dt}=8{\text{cm}}^2\text{/s}$ (note: the unit for rate of change of area is cm²/s)
- The length at the instant, $l=4\text{cm}$
- The width at the instant, $w=1\text{cm}$

Substituting these values into our differentiated equation:
$$8=4\cdot \frac{dw}{dt}+1\cdot 4$$

Simplify the equation to solve for $\frac{dw}{dt}$:
$$8=4\cdot \frac{dw}{dt}+4$$
Subtract 4 from both sides:
$$4=4\cdot \frac{dw}{dt}$$
Divide by 4:
$$\frac{dw}{dt}=1\text{cm/s}$$

Thus, the rate of change of the width is 1 cm/s.