Detailed Solution

Topic: Application of Derivatives (Optimization)

(I) SURFACE AREA IN TERMS OF X AND V

1. The volume is \( V = x^2 y \). Solving for height gives: \( y = \frac{V}{x^2} \).

2. The surface area of a closed box is \( S = 2x^2 + 4xy \).

3. Substitute \( y \):

\[ S = 2x^2 + 4x \left( \frac{V}{x^2} \right) \]

\( S = 2x^2 + \frac{4V}{x} \)

(II) FINDING THE DERIVATIVE

Differentiating \( S \) with respect to \( x \):

\[ \frac{dS}{dx} = 4x – \frac{4V}{x^2} \]

(III)(A) RELATION FOR MINIMUM SURFACE AREA

For minimum \( S \), set the derivative to zero:

\( 4x – \frac{4V}{x^2} = 0 \)   ⇒   \( 4x = \frac{4V}{x^2} \)   ⇒   \( V = x^3 \)

Now, substitute the volume formula \( V = x^2 y \):

\( x^2 y = x^3 \)   ⇒   \( y = x \)

Conclusion: Surface area is minimum when the height equals the side of the base (the box becomes a cube).

⚠️ Silly Mistake Audit: The “Closed Box” Trap

Always check if the box is open or closed! The problem explicitly states “closed from all sides,” which means your surface area formula must include two bases (\( 2x^2 \)). If you only use \( x^2 \) for an open base, your final relation (\( y=x \)) will be wrong.

(III)(B) PROOF FOR MAXIMUM VOLUME

1. Given Volume Function:

\[ V = \frac{1}{4}(Sx – 2x^3) \]

2. Find the First Derivative:

\[ \frac{dV}{dx} = \frac{1}{4}(S – 6x^2) \]

3. Set Derivative to Zero for Maxima:

\( S – 6x^2 = 0 \)   ⇒   \( 6x^2 = S \)   ⇒   \( x = \sqrt{\frac{S}{6}} \)

4. Verification using Second Derivative:

\[ \frac{d^2V}{dx^2} = \frac{1}{4}(0 – 12x) = -3x \]

Since \( x \) represents length, it is always positive. Therefore, \( -3x < 0 \).

Conclusion: By the Second Derivative Test, Volume (V) is Maximum at \( x = \sqrt{\frac{S}{6}} \).

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