Question

If the length, breadth and height of a cube are de-
creased, decreased and increased by 5%, 5% and
20%, respectively, then what will be the impact on
the surface area of the cube (in percentage terms)

Answer 1

Step-by-step explanation:

The surface area S of a cuboid with length as l , breadth b and height h is given as:

S=2⋅(lb+bh+hl)

[ 0 ]

Now, let one side of the cube C1 be x units. Its surface area S1 can be calculated as:

S1=2⋅(x⋅x+x⋅x+x⋅x)

=2⋅3x2

=6x2

[ 1 ]

Now, as given in the question statement, let l′ , b′ and h′ be the dimensions of the cuboid C2 obtained after modifying the given cube.

The surface area S2 of this solid can be calculated as:

S2=2⋅(l′b′+b′h′+h′l′)

[ 2 ]

And now comes the fun part, the percentages with which the dimensions of the cube C1 are being altered. So let’s look into that.

l′=x−5% of x=x−5100x=95100x

b′=x−5% of x=95100x

h′=x+20% of x=x+20100x=120100x

Putting these values back in 2 , we get the value of S2 as:

S2=2⋅(95100x⋅95100x+95100x⋅120100x+120100x⋅95100x)

=2⋅x2100⋅100(95⋅95+2⋅120⋅95)

=2⋅x2100⋅100(9025+22800)

=6365010000x2

[ 3 ]

This will be the new surface area of the resulting solid.

Now to find out the percentage difference between the two surface areas S1 and S2 , denoted by δ .

δ=100⋅S2−S1S1

Substituting the values from 3 and 1 , we get:

δ=100⋅6365010000x2−6x26x2

=100⋅63650−6000010000x26x2

=100⋅365060000

≈6.0834%

Or otherwise simply consider a unit cube. It will be a little easier to arrive at the solution. Whatever floats your boat.

Hope that helped.