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, $\frac{dl}{dt}$ = 8 cm/s &
$\frac{dw}{dt}$ = 3 cm/s

Area of the rectangle, a = l × w

Differentiate the equation with respect to t,

$\frac{da}{dt}$ = l × $\frac{dw}{dt}$ +  w × $\frac{dl}{dt}$

$\frac{da}{dt}$ = 20 × 3 +  10 × 8

$\frac{da}{dt}$ = 60 +  80

$\frac{da}{dt}$ = 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