Question

The width of a rectangle is increasing at a rate of 2 cm/sec, while the length increases at 3 cm/sec. At what rate is the area increasing when w = 4cm and l = 5cm?

Answer 1

Answer:

The area of the rectangle is increasing at a rate of  $22\ cm^2/s$.

Step-by-step explanation:

Given : The width of a rectangle is increasing at a rate of 2 cm/ sec. While the length increases at 3 cm/sec.

To find : At what rate is the area increasing when w = 4 cm and I = 5 cm?

Solution :

The area of the rectangle with length 'l' and width 'w' is given by  $A=l w$

Derivative w.r.t  't',

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

Now, we have given

$\frac{dl}{dt}=3\ cm/s$

$l=5\ cm$

$\frac{dw}{dt}=2\ cm/s$

$w=4\ cm$

Substitute all the values,

$\frac{dA}{dt}=(4)(3)+(5)(2)$

$\frac{dA}{dt}=12+10$

$\frac{dA}{dt}=22\ cm^2/s$

Therefore, the area of the rectangle is increasing at a rate of  $22\ cm^2/s$.

Answer 2

The area of rectangle is increasing at rate of 22 cm/ second.

Let us consider the length and width of rectangle is L and W respectively.

Given that,  $\frac{dW}{dt}=2cm/s,\frac{dL}{dt}=3cm/s$

Area of rectangle is,

                                   $A = L *W$

Differentiate above expression with respect to time t.

     $\frac{dA}{dt} =L\frac{dW}{dt}+W\frac{dL}{dt} \\\\\frac{dA}{dt}=2L+3W$

substituting w = 4cm and L = 5cm in above expression.

   $\frac{dA}{dt} =2(5)+3(4)=22cm/s$

Thus, The area of rectangle is increasing at rate of 22 cm/ second.

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