If the volume of the cube is increasing at a rate of 24 in^3 per min.?

Question

and the surface area of the cube is increasing at 12in^2 per minute. what is the length of each edge of the cube?
A) 2 in
B) 2√(2) in
C) ³√(12) in
D) 4 in
E) 8 in

Answer 1

V = s³

Differentiate:

V' = 3s² s'

We've got V' but not s'...so take a look at the surface area:

A = 6s²

Differentiate:

A' = 12s s'

Plug in the value we have for A':

12 = 12s s'

Solve for s':

s' = 1/s

Plug in the V' equation:

V' = 3s²(1/s) =3s

Plug in the value we have for V':

24 = 3s

Solve for s:

s = 8

Answer 2

Let x be the side length, v be the volume, and s be the surface area of the cube.

v = x^3
dv/dt = 3x^2 dx/dt = 24......(1)

s = 6x^2
ds/dt = 12 x dx/dt = 12......(2)

(1)/(2): x = 8 in

Answer 3

B