The surface area S of a cube with volume V is S=6V^1/3 . What effect does increasing the volume of a cube by a factor of 7 have on the on the surface area?

Question

 a.The surface area increases by a factor of . 7^2/3
b. The surface area increases by a factor of 7.
c. The surface area increases by a factor of .7^2
d.The surface area increases by a factor of .7^-1/3

Answer 1

 

First, recall the formula for calculating surface area of a cube with a side length of  s.

Each face has an area of s² and there are six faces, so the surface area A = 6s².

 

The volume V of a cube is also related only to the side length s,  V = s³.

 

When the original volume V is increased to 7V,

the side length changes to a new side length n to produce a new volume N:  

 

Using the original notation, V =7s³

or with the new notation, N = n³ where  n = (³√7)(s).

 

The new surface area N = 6(³√7(s))²  = 6s²(7)^2/3

The ratio of the new surface area to the old surface area is N/A =[ (6s²(7)^2/3 ] ^÷ 6s²

The 6s² cancels from the both the numerator and denominator,

so the ratio of the change in surface area is (7)^2/3.

This looks like answer (a) if you move the period to the end of the sentence. 

 

If (f) is the factor (f) of the change in volume, then the change in surface area is ³√(f²)  

 

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To check the above answer, it is often useful to create another problem.

Assume a cube has a side length which is originally 4 and then changes to 8.

 

Using both formulae, the surface area changes from A = 6(4)² = 96  to  N = 6(8)² = 6(64) = 384.

The ratio of change in volume of the cube is from V = 4³ to V =8³, which is from V = 64 to V = 512.

 

The factor of the change in volume is 512/64 = 8 and the surface area changes by a factor of 384/96 = 4.

This shows that the change in surface area is 3√(8²) = 3√(8²) = 3√(64) = 4.