Limited Area for a Rectangle

Calculus Level 2

The area of a rectangle is 100 square centimeters. If the length increases at the rate of 2 cm per second, find the rate of change of the width of the rectangle at the instant when the length is 4 cm.

(Note: A negative answer means that the dimension decreases at that certain rate.)

2/25 25/2 -25/2 -2/25

Roi Solomon Labay by Roi Solomon Labay , 24, Philippines

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1 solution

Jeric Garrido
May 13, 2014

Le l and w be the length and width of the rectangle, respectively. Then if the area of the rectangle is 100 c m 2 100 cm^{2} ,
⇒ l w = 100 \Rightarrow lw=100 ⇒ l = 100 w \Rightarrow l=\frac {100} {w} Also, note that since the rate of change of width is 2 c m / s 2 cm/s , we say that d w d t = 2. \frac {dw} {dt} =2. . By Chain rule, d l d t = d l d w d w d t \frac {dl} {dt} = \frac {dl} {dw} \frac {dw} {dt} But, if l = 100 / w l=100/w ⇒ d l d t = − 100 w 2 \Rightarrow \frac {dl} {dt} = - \frac {100} {w^{2}} ⇒ d l d t \arrowvert t = 2 = − 100 4 2 ( 2 ) = − 25 / 2. \Rightarrow \frac {dl} {dt} \arrowvert _{t=2} = - \frac {100} {4^{2}} (2)= -25/2.

The answer is sensible since if the area is constant, increasing the length yields a decrease in the width of the rectangle, and vice-versa.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

It should be that if l = 100 w l=\frac{100}{w} then d l d t = − 100 w 2 d w d t \frac{\text{d}l}{\text{d}t}=-\frac{100}{w^2}\frac{\text{d}w}{\text{d}t} so that d w d t = − w 2 100 d l d t \frac{\text{d}w}{\text{d}t}=-\frac{w^2}{100}\frac{\text{d}l}{\text{d}t} For a rectangle of area 100 and length 4, the width is w = 25 w=25 . Thus d w d t = − ( 25 ) 2 100 ( 2 ) = − 25 2 \frac{\text{d}w}{\text{d}t}=-\frac{(25)^2}{100}(2)=-\frac{25}{2} The above solution is still sensible, if you have used w = 100 l w=\frac{100}{l} . :)

Jaydee Lucero - 7 years, 1 month ago

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