Question

The length of a rectangle is increasing at a rate of 8cm/s8cm/s and its width is increasing at a rate of 5cm/s5cm/s. When the length is 20cm20cm and the width is 25cm,25cm, how fast is the area of the rectangle increasing

Answer 1

Answer:

The area of the given rectangle increasing when l=20cm and w=25 cm by fast is $300cm^2 per s$

Step-by-step explanation:

Given that the length of a rectangle is increasing at a rate of 8cm per s and its width is increasing at a rate of 5cm per s.

To find the  how fast is the area of the rectangle increasing when the length is 20 cm and the width is 25 cm:

Let l be the Length of Rectangle (cm)

Let w  be the Width of Rectangle (cm)

Let A  be the Area of Rectangle ($cm^2$)

Let t  be the Time (s)

From the given we can write

cm per s and  

cm per s

The formula for Area of the rectangle is:

A=lw  square units

Differentiating with respect to t

$\frac{dA}{dt}=(l)(\frac{dw}{dt})+(\frac{dl}{dt})(w)$  ( by using the product rule formula $\frac{d(uv)}{dx}=u(\frac{dv}{dx})+\frac{du}{dx}(v)$)

$=l(5)+8(w)$

when l=20 and w=25

$=(20)(5)+8(25)$

$=100+200$

$=300$

∴  $\frac{dA}{dx}=300 cm^2 per s$

∴ the area of the rectangle increasing by fast is $300cm^2 per s$