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Mathematics

The volume of a cuboid of dimensions x, y and z is V and its surface area is S. The value of 1V\dfrac{1}{V} is :

  1. 2S(1x+1y+1z)\dfrac{2}{S}\Big(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}\Big)

  2. S2(1x+1y+1z)\dfrac{S}{2} \Big(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}\Big)

  3. 3S(1x−1y−1z)\dfrac{3}{S} \Big(\dfrac{1}{x} - \dfrac{1}{y} - \dfrac{1}{z}\Big)

  4. S3(a+b+cabc)\dfrac{S}{3} \Big(\dfrac{a + b + c}{abc}\Big)

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Answer

Given,

Volume of cuboid = V

Surface area of cuboid = S

Dimensions of cuboid = x, y, z

Volume of cuboid = length × breadth × height

V = xyz

Addition of the reciprocal values of the given dimensions,

1x+1y+1z\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}

⇒ 1x+1y+1z\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = yz+xz+xyxyz\dfrac{yz + xz + xy}{xyz}

By substituting the value V = xyz, we get

⇒ 1x+1y+1z\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = yz+xz+xyV\dfrac{yz + xz + xy}{V}

⇒ V(1x+1y+1z)\Big(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}\Big) = yz + xz + xy ……….(1)

We know that,

Surface area of cuboid = 2 (lb + bh + hl)

S = 2(xy + yz + zx)

⇒ S2\dfrac{S}{2} = xy + yz + zx

By substituting the value of xy + yz + zx in equation (1), we get :

⇒S2=V(1x+1y+1z)⇒1V=2S(1x+1y+1z).\Rightarrow \dfrac{S}{2} = V\Big(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}\Big) \\[1em] \Rightarrow \dfrac{1}{V} = \dfrac{2}{S}\Big(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}\Big).

Hence, option 1 is the correct option.

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