Question

The length of a rectangle is increasing at a rate of 8 cm/s and its width is increasing at a rate of 3 cm/s. when the length is 20 cm and the width is 10cm, how fast is the area of the rectangle increasing

Answer 1

Let l = length of the rectangle,
b = breadth of the rectangle,
a = area of the rectangle,
t = time,

So,
`(dl)/(dt)` = 8 cm/s &

`(dw)/(dt)` = 3 cm/s

Area of the rectangle, a = l × w

Differentiate the equation with respect to t,


`(da)/(dt)` = l ×
`(dw)/(dt)` + w ×
`(dl)/(dt)`


`(da)/(dt)` = 20 × 3 + 10 × 8


`(da)/(dt)` = 60 + 80


`(da)/(dt)` = 140
`cm^(2)`/s

Therefore, the correct answer is 140
`cm^(2)`/s.

Answer 2

The information that is given:
dl/dt = 8 (rate of length increasing)
dw/dt = 3 (rate of width increasing)
l = 20 (when the length is 20)
w = 10 (when the width is 10)

The unknown is dA/dt (the rate the area is increasing)

Start with the formula for the area of a rectangle
A = w*l
Find the derivative of it
d/dt(A) = d/dt(w*l) Use the multiplication rule
dA/dt = w(dl/dt) + l(dw/dt)
Substitute in what is known from above
dA/dt = 10(8) + 20(3) = 80+60 = 140
The area is increasing 140 square centimeters per second