Question

The length of a rectangle is increasing at a rate of 6 cm/s and its width is increasing at a rate of 5 cm/s. When the length is 12 cm and the width is 4 cm, how fast is the area of the rectangle increasing?

Answer 1

Answer:

Area of the rectangle is increasing with the rate of 84 cm/s.

Step-by-step explanation:

Let l represents the length, w represents width, t represents time ( in seconds ) and A represents the area of the triangle,

Given,

$\frac{dl}{dt}=6\text{ cm per second}$

$\frac{dw}{dt}=5\text{ cm per second}$

Also, l = 12 cm and w = 4 cm,

We know that,

A = l × w,

Differentiating with respect to t,

$\frac{dA}{dt}=\frac{d}{dt}(l\times w)$

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

By substituting the values,

$\frac{dA}{dt}=12\times 5+4\times 6$

$=60+24$

$=84$

Hence, the area of the rectangle is increasing with the rate of 84 cm/s.