2.10 Q-7

Question Statement

A box with a square base and open top is to have a volume of $4 , \text{dm}^3$. Find the dimensions of the box that will require the least material (i.e., minimize the surface area).

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Background and Explanation

In this problem, we are asked to minimize the surface area of a box with a square base and an open top, while maintaining a fixed volume. We need to use the relationship between the box’s volume and surface area to set up an optimization problem. The volume constraint will help us express the height of the box in terms of the base length, and the surface area formula will allow us to apply calculus to minimize the material required.

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Solution

1. Define the variables:

   Let the side length of the square base be $x$, and the height of the box be $h$. The volume $V$ of the box is given by:

$$V = x^2 \cdot h = 4$$

Solving for $h$, we get:

$$h = \frac{4}{x^2} \quad \text{(Equation 1)}$$

2. Surface area function:

   The surface area $S$ of the box includes the area of the square base ($x^2$) and the areas of the four sides ($4xh$). Thus, the surface area is:

$$S = x^2 + 4xh$$

Substituting $h$ from Equation 1:

$$S = x^2 + 4x \left( \frac{4}{x^2} \right) = x^2 + \frac{16}{x}$$

3. Differentiate the surface area function:

   To minimize the surface area, we differentiate $S$ with respect to $x$:

$$\frac{dS}{dx} = \frac{d}{dx} \left( x^2 + \frac{16}{x} \right) = 2x - \frac{16}{x^2}$$

Simplifying:

$$\frac{dS}{dx} = 2 \left( \frac{x^3 - 8}{x^2} \right)$$

4. Set the derivative equal to zero:

   To find the critical points, set $\frac{dS}{dx} = 0$:

$$2 \left( \frac{x^3 - 8}{x^2} \right) = 0$$

Solving for $x$:

$$x^3 - 8 = 0 \quad \Rightarrow \quad x = 2$$

5. Check the second derivative:

   To confirm that this critical point corresponds to a minimum, compute the second derivative:

$$\frac{d^2S}{dx^2} = \frac{d}{dx} \left( 2x - \frac{16}{x^2} \right) = 2 + \frac{32}{x^3}$$

Substituting $x = 2$:

$$\frac{d^2S}{dx^2} = 2 + \frac{32}{2^3} = 2 + \frac{32}{8} = 2 + 4 = 6$$

Since the second derivative is positive, $x = 2$ corresponds to a minimum.

6. Find the height $h$:

   Now substitute $x = 2$ into Equation 1 to find the height:

$$h = \frac{4}{2^2} = \frac{4}{4} = 1$$

Thus, the dimensions of the box that will require the least material are 2 dm by 2 dm for the base, and 1 dm for the height.

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Key Formulas or Methods Used

• Volume of the box: $V = x^2 \cdot h$
• Surface area of the box: $S = x^2 + 4xh$
• First derivative test: To find the critical points and minimize the surface area.
• Second derivative test: To confirm that the critical point corresponds to a minimum.

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Summary of Steps

1. Express the height $h$ in terms of $x$ using the volume equation: $h = \frac{4}{x^2}$.
2. Set up the surface area function: $S = x^2 + \frac{16}{x}$.
3. Differentiate the surface area function: $\frac{dS}{dx} = 2 \left( \frac{x^3 - 8}{x^2} \right)$.
4. Set $\frac{dS}{dx} = 0$ and solve for $x = 2$.
5. Confirm the minimum using the second derivative test.
6. Find the height: $h = 1$ dm.
7. The dimensions of the box are 2 dm by 2 dm for the base, and 1 dm for the height.