A covered box of volume `72 cm^(3)` and the base sides in a ratio of `1: 2` is to be made. The length all sides so that the total surface area is the least possible is

Options

A. 2,4,9
B. 8,3,3
C. 6,6,2
D. 6,3,4

Answer

To solve the problem of finding the dimensions of a covered box with a volume of 72 cm³ and base sides in a ratio of 1:2 that minimizes the total surface area, we can follow these steps:

### Step 1: Define Variables
Let the length of the base be \( x \) and the breadth be \( 2x \). Let the height of the box be \( y \).

### Step 2: Write the Volume Equation
The volume \( V \) of the box is given by:
\[ V = \text{length} \times \text{breadth} \times \text{height} = x \times 2x \times y = 2x^2y \]
Given that the volume is \( 72 \) cm³, we have:
\[ 2x^2y = 72 \]
From this, we can express \( y \) in terms of \( x \):
\[ y = \frac{72}{2x^2} = \frac{36}{x^2} \]

### Step 3: Write the Total Surface Area Equation
The total surface area \( S \) of the box is given by:
\[ S = 2(\text{length} \times \text{breadth}) + 2(\text{length} \times \text{height}) + 2(\text{breadth} \times \text{height}) \]
Substituting the values, we get:
\[ S = 2(x \times 2x) + 2(x \times y) + 2(2x \times y) \]
This simplifies to:
\[ S = 4x^2 + 2xy + 4xy = 4x^2 + 6xy \]
Now substituting \( y \):
\[ S = 4x^2 + 6x\left(\frac{36}{x^2}\right) = 4x^2 + \frac{216}{x} \]

### Step 4: Differentiate the Surface Area
To find the minimum surface area, we need to differentiate \( S \) with respect to \( x \):
\[ \frac{dS}{dx} = \frac{d}{dx}(4x^2) + \frac{d}{dx}\left(\frac{216}{x}\right) \]
Calculating the derivatives:
\[ \frac{dS}{dx} = 8x - \frac{216}{x^2} \]

### Step 5: Set the Derivative to Zero
To find the critical points, set the derivative equal to zero:
\[ 8x - \frac{216}{x^2} = 0 \]
Multiplying through by \( x^2 \) to eliminate the fraction:
\[ 8x^3 - 216 = 0 \]
This simplifies to:
\[ 8x^3 = 216 \quad \Rightarrow \quad x^3 = \frac{216}{8} = 27 \quad \Rightarrow \quad x = 3 \]

### Step 6: Find the Corresponding Dimensions
Now that we have \( x = 3 \):
- The breadth \( b = 2x = 2 \times 3 = 6 \) cm
- The height \( y = \frac{36}{x^2} = \frac{36}{3^2} = \frac{36}{9} = 4 \) cm

### Final Dimensions
Thus, the dimensions of the box are:
- Length \( x = 3 \) cm
- Breadth \( 2x = 6 \) cm
- Height \( y = 4 \) cm

### Summary of Steps
1. Define variables for length, breadth, and height.
2. Write the volume equation and solve for height in terms of length.
3. Write the total surface area equation and substitute for height.
4. Differentiate the surface area with respect to length.
5. Set the derivative equal to zero to find critical points.
6. Solve for dimensions using the critical point.