Final answer:
The area of the rectangle is increasing at a rate of 200 cm²/s when the length is 40 cm and the width is 20 cm. This is calculated by applying the product rule to the rate of change of the area in relation to time.
Step-by-step explanation:
The question is asking how fast the area of a rectangle is increasing given the rates at which its length and width are increasing. To find out how fast the area is increasing, we will be using the concept of related rates in calculus.
Given that the length (L) is increasing at 6 cm/s and the width (W) is increasing at 2 cm/s, when the length is 40 cm and the width is 20 cm, we can set up the equation for the rate of change of the area (A) using the formula A = L x W.
The rate of change of the area, also known as the derivative of the area with respect to time (dA/dt), is given by the product rule of differentiation:
dA/dt = L'(t) x W + L x W'(t)
Where:
- L'(t) = rate of change of length = 6 cm/s
- W'(t) = rate of change of width = 2 cm/s
Substituting the values:
dA/dt = (6 cm/s x 20 cm) + (40 cm x 2 cm/s)
dA/dt = 120 cm²/s + 80 cm²/s
dA/dt = 200 cm²/s
Thus, the area of the rectangle is increasing at a rate of 200 cm²/s when the length is 40 cm and the width is 20 cm.