Question

The length $x$ of a rectangle is increasing at $5\;{\text{cm/s}}$ and breadth is decreasing at $4\;{\text{cm/s}}$, when $x = 24\;{\text{cm}}$ and $y = 20\;{\text{cm}}$. Find the rate of change of perimeter and area of the rectangle.

Answer 1

Hint: We know that the rate of change in length of the rectangle is represented by positive sign as it is increasing and the rate of change in breadth is represented by negative sign as it is decreasing. Put the values of rate in formulas of perimeter and area of rectangle.

Complete step-by-step answer:
We know that the rate of change of $y$ with respect to $x$ is given by $\dfrac{{dy}}{{dx}}$.
Let us consider $x$ be the length and $y$ be the breadth of rectangle respectively.
 The rate of increasing length is $5\;{\text{cm/s}}$. We can say that $x$ is increasing with respect to time.
\[\dfrac{{dx}}{{dt}} = 5\;{\text{cm/s}}\]
 The rate of decreasing breadth is $4\;{\text{cm/s}}$. We can say that $y$ is decreasing with respect to time.
\[\dfrac{{dy}}{{dt}} = - 4\;{\text{cm/s}}\]
 We have to calculate the rate of change of perimeter when $x = 24\;{\text{cm}}$ and $y = 20\;{\text{cm}}$. The formula for perimeter $P$ of the rectangle is:
$P = 2\left( {x + y} \right)$
 Now, differentiate the perimeter with respect to time.
$ \Rightarrow \dfrac{{dP}}{{dt}} = \dfrac{{d\left[ {2\left( {x + y} \right)} \right]}}{{dt}}$
$ \Rightarrow \dfrac{{dP}}{{dt}} = \dfrac{{d\left( {2x} \right)}}{{dt}} + \dfrac{{d\left( {2y} \right)}}{{dt}}$
$ \Rightarrow \dfrac{{dP}}{{dt}} = 2\dfrac{{dx}}{{dt}} + 2\dfrac{{dy}}{{dt}}$
 Substitute the values and solve.
$ \Rightarrow \dfrac{{dP}}{{dt}} = 2\left( {5\;{\text{cm/s}}} \right) + 2\left( { - 4\;{\text{cm/s}}} \right)$
$ \Rightarrow \dfrac{{dP}}{{dt}} = 10\;{\text{cm/s}} - 8\;{\text{cm/s}}$
$ \Rightarrow \dfrac{{dP}}{{dt}} = 2\;{\text{cm/s}}$
So, the perimeter is increasing at a rate of $2\;{\text{cm/s}}$.
We have to calculate the rate of change of perimeter when $x = 24\;{\text{cm}}$ and $y = 20\;{\text{cm}}$. The formula for area $A$ of the rectangle is:
$A = xy$
Now, differentiate the area with respect to time.
$ \Rightarrow \dfrac{{dA}}{{dt}} = \dfrac{{d\left[ {xy} \right]}}{{dt}}$
$ \Rightarrow \dfrac{{dA}}{{dt}} = x\dfrac{{dy}}{{dt}} + y\dfrac{{dx}}{{dt}}$
 Substitute the values and solve.
\[ \Rightarrow \dfrac{{dA}}{{dt}} = \left( {24\;{\text{cm}}} \right)\left( { - 4\;{\text{cm/s}}} \right) + \left( {20\;{\text{cm}}} \right)\left( {5\;{\text{cm/s}}} \right)\]
\[ \Rightarrow \dfrac{{dA}}{{dt}} = - 96\;{\text{c}}{{\text{m}}^2}{\text{/s}} + 100\;{\text{c}}{{\text{m}}^2}{\text{/s}}\]
\[ \Rightarrow \dfrac{{dA}}{{dt}} = 4\;{\text{c}}{{\text{m}}^2}{\text{/s}}\]
So, the area is increasing at a rate of $4\;{\text{c}}{{\text{m}}^2}{\text{/s}}$.
Therefore, the perimeter of rectangle is increasing at the rate of $2\;{\text{cm/s}}$ and area is increasing at rate of $4\;{\text{c}}{{\text{m}}^2}{\text{/s}}$.

Note: We use positive signs to represent the increasing rate of a quantity and negative sign to represent the decreasing rate of a quantity. Using the opposite signs for the rate of change may lead to the incorrect answer.