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# 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 rates of change of (b) the area of the triangle.

This is (b) part of the multi-part question q7

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:
The rate of change in length is $\large\frac{dx}{dt}$$=-5cm/min. Negative sign indicates that the length is decreasing. The rate of change in width is \large\frac{dy}{dt}=$$4cm/min$
Also it is given that the length and width are $x=8cm$ and $y=6cm$ respectively.
Area of the triangle is $xy$
$A=xy$
Step 2:
Differentiate on both sides w.r.t $t$
(Apply product rule)
$\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}$
Step 3:
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 $2cm^2/min$