A Rectangular Box Without A Lid Is To Be Made From 12 M Ofcardboard. Find The Maximum Volume Of Such

A Rectangular Box Without A Lid Is To Be Made From 12 M Of Cardboard. Find The Maximum Volume Of Such

Introduction

Creating a box with maximum volume using a limited amount of material is a classic optimization problem in calculus and geometry. In this particular problem, we are tasked with designing a rectangular box without a lid, using only 12 meters of cardboard. The goal is to determine the dimensions that will yield the maximum possible volume. This scenario involves understanding the relationship between surface area and volume, setting up an appropriate mathematical model, and applying calculus techniques such as differentiation to find the optimal dimensions.

Problem Restatement and Assumptions

Before delving into calculations, let's clarify the problem's parameters and assumptions:

• The box is rectangular and has no lid.
• The total cardboard used (surface area) is 12 meters.
• The box is open at the top.
• The dimensions are length (l), width (w), and height (h), all positive real numbers.
• The goal is to maximize the volume (V) of the box.

Given these assumptions, our task is to find the dimensions (l, w, h) that maximize the volume under the given surface area constraint.

Formulating the Mathematical Model

Surface Area Constraint

Since the box has no lid, the surface area (SA) comprises:

• The bottom: l × w
• The four vertical sides: 2(l × h) + 2(w × h)

Total surface area: \[ SA = l w + 2 l h + 2 w h \] Given: \[ SA = 12 \text{ meters} \] So, \[ l w + 2 l h + 2 w h = 12 \]

Volume Expression

The volume (V) of the box is: \[ V = l \times w \times h \]

Our goal is to maximize V subject to the surface area constraint.

Reducing the Problem

To optimize V, it's convenient to express it in terms of fewer variables. Since the surface area constraint involves l, w, and h, we can reduce the problem by expressing one variable in terms of the others.

A practical approach is to fix the relationship between l and w or to assume symmetry. For simplicity, assume the base is square, i.e.,

\[
l = w
\]

This is a common assumption in such problems to reduce complexity, and often the maximum volume occurs when the base is square.

Thus:
\[
l = w = x
\]

Now, the surface area becomes:
\[
SA = x \times x + 2 x h + 2 x h = x^2 + 4 x h
\]
and the volume:
\[
V = x \times x \times h = x^2 h
\]

Given that:
\[
x^2 + 4 x h = 12
\]
we can solve for h:
\[
4 x h = 12 - x^2
\]
\[
h = \frac{12 - x^2}{4 x}
\]

The volume function V in terms of x:
\[
V(x) = x^2 \times h = x^2 \times \frac{12 - x^2}{4 x} = \frac{x^2 (12 - x^2)}{4 x}
\]

Simplify:
\[
V(x) = \frac{x (12 - x^2)}{4}
\]
\[
V(x) = \frac{12 x - x^3}{4}
\]

Now, the domain of x is constrained by the requirement that h > 0:
\[
h > 0 \Rightarrow 12 - x^2 > 0 \Rightarrow x^2 < 12 \Rightarrow x < \sqrt{12} \approx 3.464
\]
and since x > 0, the domain is:
\[
0 < x < \sqrt{12}
\]

Optimizing the Volume

Step 1: Derivative of V(x)

Differentiate V(x) with respect to x: \[ V'(x) = \frac{1}{4} (12 - 3 x^2) \]

Step 2: Find Critical Points

Set the derivative to zero to find potential maxima: \[ V'(x) = 0 \Rightarrow 12 - 3 x^2 = 0 \] \[ 3 x^2 = 12 \] \[ x^2 = 4 \] \[ x = 2 \quad (\text{since } x > 0) \]

Step 3: Verify the Critical Point

Calculate the volume at \(x=2\): \[ h = \frac{12 - (2)^2}{4 \times 2} = \frac{12 - 4}{8} = \frac{8}{8} = 1 \] Volume: \[ V(2) = \frac{12 \times 2 - (2)^3}{4} = \frac{24 - 8}{4} = \frac{16}{4} = 4 \]

Step 4: Check the Endpoints

• As \(x \to 0^+\), volume \(V \to 0\).
• As \(x \to \sqrt{12}^-\), \(h \to 0^+\), volume approaches zero again.

Therefore, the critical point at \(x=2\) gives the maximum volume.

Final Dimensions and Maximum Volume

• Base sides:

\[ l = w = 2 \text{ meters} \]

• Height:

\[ h = 1 \text{ meter} \]

Maximum volume:
\[
V_{max} = 4 \text{ cubic meters}
\]

Conclusion

By assuming a square base, we've determined that the maximum volume of a lidless rectangular box constructed from 12 meters of cardboard occurs when the base sides are 2 meters each, and the height is 1 meter. The maximum volume achievable under these conditions is 4 cubic meters.

This problem illustrates the importance of setting up an appropriate mathematical model, simplifying assumptions to reduce complexity, and applying calculus techniques such as differentiation to solve optimization problems. Such approaches are fundamental in design, engineering, and resource management where maximizing efficiency within constraints is essential.

Extensions and Considerations

• If the problem does not assume a square base, the optimization becomes more complex and involves multiple variables.
• Real-world applications may include material thickness, structural stability, or additional constraints, which would modify the model.
• The method demonstrated here can be adapted to similar problems involving surface area and volume optimization under different conditions.