if the length is decreasing at a rate of 2 in./min. while width is increasing at a rate of 2 in./min., what must be true about area of the rectangle
A = Lw
dA/dt = L dw/dt + w dL/dt
but dL/dt = -2
and dw/dt = 2
dA/dt = 2L - 2w
= 2(L-w)
So it depends on the size of the rectangle,
If the rectangle is a square (L=w), then area would remain constant,
if L > w, the area is increasing, and
if L < w ......
To determine the relationship between the length, width, and area of a rectangle, let's use the formula for the area of a rectangle: A = l * w, where A represents the area, l represents the length, and w represents the width.
Given that the length is decreasing at a rate of 2 in./min, this can be represented as dl/dt = -2 in./min (negative because it is decreasing).
Similarly, the width is increasing at a rate of 2 in./min, which can be represented as dw/dt = 2 in./min.
To find the rate of change of the area, we can take the derivative of the area formula with respect to time (t). Let's differentiate both sides of the equation A = l * w with respect to t:
dA/dt = d/dt (l * w)
Using the product rule for differentiation, the derivative of a product is the first term times the derivative of the second term, plus the second term times the derivative of the first term:
dA/dt = l * dw/dt + w * dl/dt
Substituting the given rates of change, we have:
dA/dt = l * 2 + w * (-2)
Simplifying further:
dA/dt = 2l - 2w
Therefore, the rate of change of the area of the rectangle is given by dA/dt = 2l - 2w.
In this case, the rate of change of the area depends on the difference between twice the length and twice the width. Depending on the values of length and width, the area may increase, decrease, or remain constant.