Question

The length of a rectangle is increasing at a rate of 4 cm/s and its width is increasing at a rate of 8 cm/s. When the length is 8 cm and the width is 5 cm, how fast is the area of the rectangle increasing?

Answer 1

Answer:

Step-by-step explanation:

$A = lw \nl\frac{\dA}{\dt}$ =$\frac{\dA}{\text{d}l}\frac{\text{d}l}{\dt} + \frac{\dA}{\text{d}w}\frac{\text{d}w}{\dt}$ =$w\frac{\text{d}l}{\dt} + l\frac{\text{d}w}{\dt}$ $\nll = 20 \text{ cm}$ \;\; $\frac{\text{d}l}{\dt} = 8 \text{ cm/s}$ \;\;w = 10 \text{ cm}, \;\;  $]\frac{\text{d}w}{\dt} = 3 \text{ cm/s}$ $\nl\frac{\dA}{\dt} =(  10 \text{ cm} )( 8 \text{ cm/s} ) + (  20 \text{ cm} )( 3 \text{ cm/s} )  =140 \text{ cm}^2\!\text{/s}$

Answer 2

Given :

• The length of a rectangle is increasing at a rate of 4 cm/s and its width is increasing at a rate of 8 cm/s. When the length is 8 cm and the width is 5 cm .

To find :-

• how fast is the area of the rectangle increasing?

Solution :-

As we know that :-

• A = lb

To find the rate :-

• d(A)/dt = d(lb)/dt .

Differenciate :-

• dA/dt = l (db/dt ) + b (dl/dt )

Substitute :-

• dA/dt = 8*8 + 5*4
• dA/dt = 64 + 20 cm²/s
• dA/dt = 84 cm²/s