Browse Questions

# The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rate of change of (a) the perimeter, (b) the area of the rectangle.

Toolbox:
• If $y=f(x)$,then $\large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $x$.
• $\big(\large\frac{dy}{dx}\big)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$
Step 1:
Rate of decrease in length is $\large\frac{dx}{dt}$$=-5cm/min. Negative sign shows it is decreasing. Rate of increase in width is \large\frac{dy}{dt}$$=4cm/min$.
Perimeter of the rectangle is $P=2x+2y$
Differentiating w.r.t $t$ on both sides we get,
$\large\frac{dp}{dt}=$$2.\large\frac{dx}{dt}+$$2.\large\frac{dy}{dt}$
Step 2:
Now substituting the values for $\large\frac{dx}{dt}$ and $\large\frac{dy}{dt}$ we get,
$\large\frac{dp}{dt}=$$2(-5)+2(4) \quad\;\;=-2cm/min. Hence the perimeter decreases at the rate of -2cm/min. Step 3: Area of the rectangle is xy A=xy Differentiate on both sides w.r.t t \large\frac{d}{dt}$$(uv)=v\large\frac{du}{dt}+$$u.\large\frac{dv}{dt} \large\frac{dA}{dt}=$$x.\large\frac{dy}{dt}+$$y.\large\frac{dx}{dt} Substituting the values for x,y,\large\frac{dy}{dt} and \large\frac{dx}{dt} is \large\frac{dA}{dt}=$$8\times (4)+4\times (-5)$
$\quad\;\;\;=32-30$
$\quad\;\;\;=2cm^2/min$
Hence the area of the rectangle increases at the rate of $2cm^2/min$