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
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
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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: \( \frac{dV}{dt} = \frac{x}{4} \cdot \frac{dS}{dt} \).
Updated On: Jun 8, 2026
- \( 6 \)
- \( 12 \)
- \( 18 \)
- \( 9 \)
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The Correct Option is C
Solution and Explanation
Concept:
Let \( x \) be the side length of the cube, \( S \) be its total surface area, and \( V \) be its volume:
\[
S = 6x^2, \quad V = x^3
\]
We use the chain rule to relate their rates of change with respect to time \( t \).
Step 1: Finding the side rate of change \( \frac{dx}{dt} \).
Differentiating the surface area formula with respect to time: \[ \frac{dS}{dt} = 12x \frac{dx}{dt} \] We are given \( \frac{dS}{dt} = 6 \) and the current side length \( x = 12 \) cm: \[ 6 = 12(12) \frac{dx}{dt} \implies 6 = 144 \frac{dx}{dt} \implies \frac{dx}{dt} = \frac{6}{144} = \frac{1}{24} \text{ cm/sec} \]
Step 2: Calculating the rate of change of volume.
Differentiating the volume formula with respect to time: \[ \frac{dV}{dt} = 3x^2 \frac{dx}{dt} \] Substitute \( x = 12 \) and \( \frac{dx}{dt} = \frac{1}{24} \): \[ \frac{dV}{dt} = 3(12)^2 \left(\frac{1}{24}\right) = 3(144) \left(\frac{1}{24}\right) = 3 \times 6 = 18 \text{ c.c./sec} \] Let us re-verify option formatting keys. Under standardized base conversions, the calculation yields 18, matching option (C) perfectly.
Step 1: Finding the side rate of change \( \frac{dx}{dt} \).
Differentiating the surface area formula with respect to time: \[ \frac{dS}{dt} = 12x \frac{dx}{dt} \] We are given \( \frac{dS}{dt} = 6 \) and the current side length \( x = 12 \) cm: \[ 6 = 12(12) \frac{dx}{dt} \implies 6 = 144 \frac{dx}{dt} \implies \frac{dx}{dt} = \frac{6}{144} = \frac{1}{24} \text{ cm/sec} \]
Step 2: Calculating the rate of change of volume.
Differentiating the volume formula with respect to time: \[ \frac{dV}{dt} = 3x^2 \frac{dx}{dt} \] Substitute \( x = 12 \) and \( \frac{dx}{dt} = \frac{1}{24} \): \[ \frac{dV}{dt} = 3(12)^2 \left(\frac{1}{24}\right) = 3(144) \left(\frac{1}{24}\right) = 3 \times 6 = 18 \text{ c.c./sec} \] Let us re-verify option formatting keys. Under standardized base conversions, the calculation yields 18, matching option (C) perfectly.
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