$\begin{array}{1 1} (A)\;-0.12x^2m^2 \\(B)\;-0.14x^2m^2 \\(C)\;-0.16x^2m^2 \\ (D)\;-0.10x^2m^2 \end{array} $

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- Let $y=f(x)$
- $\Delta x$ denote a small increment in $x$
- $\Delta y=f(x+\Delta x)-f(x)$
- $dy=\big(\large\frac{dy}{dx}\big)\Delta x$
- Surface area of cube =$6s^2$

Step 1:

The side of the cube =$x$ meters

Decrease in side =1%

$\qquad\qquad\;\;\;\;\;\;=0.01x$

Increase in side =$\Delta x$

$\qquad\qquad\;\;\;\;\;=-0.01x$

Step 2:

Surface area of cube =$6s^2$

$\qquad\qquad\qquad\;\;\;=6\times x^2$

$S= 6x^2$

$\large\frac{ds}{dx}$ = 12x [Differentiating with respect to x]

Approximate change in surface area of cube =$\large\frac{ds}{dx}$$\times \Delta x$

$\qquad\qquad\;\;\;\;\;\;\;\qquad\qquad\qquad\qquad\quad\;=12x\times (-0.01x)$

$\qquad\qquad\;\;\;\;\;\;\;\qquad\qquad\qquad\qquad\quad\;=-0.12x^2m^2$

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