Question

if V is the volume of a cuboid of dimension x,y,z and A is its surface area then A/V

Answer 1

Given, dimensions of a cuboid are x, y and z.
Now, volume of cube = lbh = xyz
and surface area of cuboid = 2(lb + bh + hl) = 2(xy + yz + zx)
Therefore,A/V=2(lb+BH+hl)/xyz
=2(1/x+1/y+1/z)

Answer 2

$\dfrac{A}{V}=2(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})$

Explanation:

Given,

The length, breadth and height of a cuboid = x, y and z.

To find, the ratio of $\dfrac{A}{V}=?$

We know that,

The volume of cuboid = Length × Breadth × Height

⇒ V = xyz      

Also, we know that

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

A = 2(xy + yz + zx)

∴ $\dfrac{A}{V}=\dfrac{2(xy + yz + zx)}{xyz}$

Dividing numerator and denoinator by xyz, we get

$\dfrac{A}{V}=\dfrac{\dfrac{2(xy + yz + zx)}{xyz}}{\dfrac{xyz}{xyz} }$

$=\dfrac{{2(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})}}{1}$

$=2(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})$

Hence, $\dfrac{A}{V}=2(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})$