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

$\begin{array}{1 1} (A)\;-2cm/min. \\(B)\;2cm/min. \\ (C)\;-3cm/min. \\ (D)\;3cm/min.\end{array} $

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

$\begin{array}{1 1} (A)\;-2cm/min. \\(B)\;2cm/min. \\ (C)\;-3cm/min. \\ (D)\;3cm/min.\end{array} $

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- 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$.

Given : $x=8cm.$ and $y=6cm$

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)$

$\quad\;\;=-2$cm/min.

Hence the perimeter decreases at the rate of $-2$cm/min.

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