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

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Side of cube decreased by 1%(\Delta x) = -0.01x m
The surface area of cube = 6a^2 \ m^2
We know that, (\Delta y) is approximately equal to dy

dy = \frac{dy}{dx}.\Delta x\\ dy = 12a(-0.01x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = 6a^2 \ and \ \Delta x = -0.01x)\\ dy = 12x(-0.01x)\\ dy=-0.12x^2 \ m^2
Hence,  the approximate change in the surface area of a cube of side x metres
caused by decreasing the side by 1%. is -0.12x^2 \ m^2

Posted by

Gautam harsolia

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