Question:
The surface area of a cube is 150 sq. cm. If it is increased by 0.025 sq. cm, then the approximate increase in its volume (in c.c.) is:
The surface area of a cube is 150 sq. cm. If it is increased by 0.025 sq. cm, then the approximate increase in its volume (in c.c.) is:
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When asked for an approximate change, use differentials (\(dV \approx V'(x)dx\)) rather than calculating the difference between two exact values. This method is much faster and accurate for small changes.
Updated On: Oct 7, 2026
- \( 0.0725 \)
- \( 0.04 \)
- \( 0.032 \)
- \( 0.03125 \)
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The Correct Option is D
Solution and Explanation
Concept:
For a cube with side length \(x\), the surface area \(S\) and volume \(V\) are given by:
\[
S = 6x^2
\]
\[
V = x^3
\]
We are looking for the approximate increase in volume (\(dV\)) given a small increase in surface area (\(dS = 0.025\)).
Step 1: Find the side length \(x\) and express \(dV\) in terms of \(dS\).
Given \(S = 6x^2 = 150\): \[ x^2 = \frac{150}{6} = 25 \] \[ x = 5 \, \text{cm} \] Now, differentiate \(S\) and \(V\) with respect to \(x\): \[ \frac{dS}{dx} = 12x \] \[ \frac{dV}{dx} = 3x^2 \]
Step 2: Relate \(dV\) and \(dS\) using the chain rule.
We can write: \[ dV = \left( \frac{dV}{dx} \right) dx \quad \text{and} \quad dS = \left( \frac{dS}{dx} \right) dx \] Thus: \[ \frac{dV}{dS} = \frac{dV/dx}{dS/dx} = \frac{3x^2}{12x} = \frac{x}{4} \]
Step 3: Calculate the approximate increase in volume.
Substitute \(x = 5\) and \(dS = 0.025\): \[ dV = \frac{x}{4} \cdot dS \] \[ dV = \frac{5}{4} \cdot 0.025 \] \[ dV = 1.25 \cdot 0.025 \] \[ dV = 0.03125 \, \text{c.c.} \]
Conclusion: The approximate increase in the volume of the cube is \( 0.03125 \, \text{c.c.} \). center Final Answer: (D) 0.03125 center
Step 1: Find the side length \(x\) and express \(dV\) in terms of \(dS\).
Given \(S = 6x^2 = 150\): \[ x^2 = \frac{150}{6} = 25 \] \[ x = 5 \, \text{cm} \] Now, differentiate \(S\) and \(V\) with respect to \(x\): \[ \frac{dS}{dx} = 12x \] \[ \frac{dV}{dx} = 3x^2 \]
Step 2: Relate \(dV\) and \(dS\) using the chain rule.
We can write: \[ dV = \left( \frac{dV}{dx} \right) dx \quad \text{and} \quad dS = \left( \frac{dS}{dx} \right) dx \] Thus: \[ \frac{dV}{dS} = \frac{dV/dx}{dS/dx} = \frac{3x^2}{12x} = \frac{x}{4} \]
Step 3: Calculate the approximate increase in volume.
Substitute \(x = 5\) and \(dS = 0.025\): \[ dV = \frac{x}{4} \cdot dS \] \[ dV = \frac{5}{4} \cdot 0.025 \] \[ dV = 1.25 \cdot 0.025 \] \[ dV = 0.03125 \, \text{c.c.} \]
Conclusion: The approximate increase in the volume of the cube is \( 0.03125 \, \text{c.c.} \). center Final Answer: (D) 0.03125 center
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