Question

If the rate of increase in the surface area of a cube is 6 sq. cm./sec., then the rate of increase in its volume (in c. c./sec), when the length of its edge is 12 cm, is

• A. \( 6 \)
• B. \( 12 \)
• C. \( 18 \)
• D. \( 9 \)

Hint:

You can find a direct relationship between the differential variables without using \( t \) explicitly: Since \( V = x^3 \) and \( S = 6x^2 \), we have \( dV = 3x^2 dx \) and \( dS = 12x dx \). Dividing them gives the clean shortcut formula: \( \frac{dV}{dt} = \frac{x}{4} \cdot \frac{dS}{dt} \).

Answer 1

The Correct Option is C

Step 1: Set up the formulas.
Let the edge be $x$. The surface area is $S=6x^2$ and the volume is $V=x^3$.
Step 2: Differentiate the surface area in time.
\[ \frac{dS}{dt}=12x\frac{dx}{dt} \]
Step 3: Find $\frac{dx}{dt}$.
We are given $\frac{dS}{dt}=6$ and $x=12$: \[ 6=12(12)\frac{dx}{dt}=144\frac{dx}{dt}\implies\frac{dx}{dt}=\frac{1}{24} \]
Step 4: Differentiate the volume in time.
\[ \frac{dV}{dt}=3x^2\frac{dx}{dt} \]
Step 5: Substitute the values.
\[ \frac{dV}{dt}=3(12)^2\cdot\frac{1}{24}=3(144)\cdot\frac{1}{24} \]
Step 6: Simplify.
\[ =3\times6=18 \] \[ \boxed{18\text{ c.c./sec}} \]