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Find the approximate change in the surface area of a cube of side \(x\) metres caused by decreasing the side by $1\%$.

$\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%
Increase in side =$\Delta x$
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$
$\qquad\qquad\;\;\;\;\;\;\;\qquad\qquad\qquad\qquad\quad\;=12x\times (-0.01x)$
answered Aug 6, 2013 by sharmaaparna1

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