IBDP Math AAHL IA: Optimizing Camping

Tent Dimensions

1. Introduction
• Research Question: What are the optimum dimensions of a square-based pyramid
camping tent to maximize internal volume with a fixed amount of material?
• Background: Briefly introduce the concept of optimization in mathematics and its
real-world applications. Discuss the practical relevance of designing efficient
camping tents.
• Aim: Clearly state the aim of the investigation: to determine the optimal
dimensions of a square-based pyramid tent under two different constraints to
maximize its internal volume.
• Scope: Outline the specific mathematical concepts and tools that will be used (e.g.,
calculus, differentiation, geometric formulas).

2. Problem 1: Maximizing Volume with Fixed Surface

Area
• Introduction to the Problem: Reiterate the first optimization problem: maximizing
the volume of a square-based pyramid with a fixed surface area (e.g., 5 m²).
• Mathematical Model:
◦ Define variables (e.g., base side length s , slant height l , vertical height h ,
volume V , surface area A ).
◦ Formulate equations for the volume and surface area of a square-based
pyramid in terms of these variables.
◦ Express volume as a function of a single variable using the surface area
constraint.
• Optimization Process:
◦ Use calculus (differentiation) to find the critical points of the volume function.
◦ Apply the second derivative test to confirm maximum volume.
◦ Calculate the optimum dimensions (s, h) and the maximum volume for the
given surface area.
• Discussion of Results: Analyze the practicality of the obtained dimensions. Discuss
why this solution might lead to an "unpractical" tent (very tall or very flat), as
mentioned by the teacher.
3. Problem 2: Maximizing Volume with Fixed Base
Hypotenuse
• Introduction to the Problem: Reiterate the second optimization problem:
maximizing the volume of a square-based pyramid with a fixed length of the base
hypotenuse (e.g., 160cm, 170cm, 180cm).
• Mathematical Model:
◦ Define variables (e.g., base side length s , slant height l , vertical height h ,
volume V , base hypotenuse d ).
◦ Formulate equations for the volume and the relationship between base side
length and base hypotenuse.
◦ Express volume as a function of a single variable using the fixed base
hypotenuse constraint.
• Optimization Process:
◦ Use calculus (differentiation) to find the critical points of the volume function.
◦ Apply the second derivative test to confirm maximum volume.
◦ Calculate the optimum dimensions (s, h) and the maximum volume for each
given base hypotenuse length.
• Discussion of Results: Compare the results for different hypotenuse lengths.
Discuss how this constraint addresses the practicality issue from Problem 1.

4. Comparative Analysis and Conclusion

• Comparison of Solutions: Compare the optimal dimensions and volumes
obtained from both problems. Discuss the trade-offs between maximizing volume
and practical design considerations.
• Limitations and Assumptions: Discuss any limitations of the mathematical
models (e.g., ideal pyramid shape, no consideration for tent poles, zippers, etc.).
• Extension/Further Research (Optional): Suggest possible extensions to the
investigation (e.g., different tent shapes, incorporating material thickness,
considering wind resistance).
• Conclusion: Summarize the key findings and answer the research question based
on the analysis of both problems.

5. Bibliography
• List all sources used (e.g., textbooks, websites, academic papers).
6. Appendix (Optional)
• Include any raw data, detailed calculations, or graphs that are too extensive for the
main body of the report.

Mathematical Details for Problem 1: Maximizing

Volume with Fixed Surface Area

Variables and Formulas:

Let: * s = side length of the square base * h = vertical height of the pyramid (from the
center of the base to the apex) * l = slant height of the triangular faces (height of each
triangular face) * V = Volume of the pyramid * A = Surface Area of the pyramid (fixed at
5 m² in your example)

Formulas: 1. Volume of a square-based pyramid: V = (1/3) * s² * h

1. Surface Area of a square-based pyramid: The surface area consists of the base
area and the area of the four triangular faces.

◦ Base Area = s²
◦ Area of one triangular face = (1/2) * base * height = (1/2) * s * l
◦ Total Area of four triangular faces = 4 * (1/2) * s * l = 2sl
◦ Therefore, A = s² + 2sl

2. Relationship between h, s, and l: Consider a right-angled triangle formed by the

vertical height h , half of the base side length ( s/2 ), and the slant height l . By
the Pythagorean theorem: l² = h² + (s/2)² l = √(h² + s²/4)

Expressing Volume as a Function of a Single Variable:

Our goal is to maximize V , but V is currently a function of s and h . We need to use the
fixed surface area constraint ( A = 5 m² ) to express V in terms of a single variable
(either s or h ). It\'s often easier to express h in terms of s and A (or l in terms of s
and A , then h in terms of s and l ).

From the surface area formula: A = s² + 2sl We can isolate l : 2sl = A - s² l =

(A - s²) / (2s)
Now substitute this expression for l into the Pythagorean relationship to find h in
terms of s and A : h² = l² - (s/2)² h² = ((A - s²) / (2s))² - (s/2)² h
= √[((A - s²) / (2s))² - (s/2)²]

Finally, substitute this expression for h into the volume formula: V(s) = (1/3) * s²
* √[((A - s²) / (2s))² - (s/2)²]

This is the function you need to differentiate. Remember to substitute A = 5 into the
equation before differentiating.

Optimization Process (using Calculus):

1. Simplify V(s): Before differentiating, it\"s often helpful to simplify the expression
for V(s) as much as possible. We will work with V²(s) to avoid the square root,
as maximizing V²(s) is equivalent to maximizing V(s) since V(s) is always
positive.

We have V(s) = (1/3) * s² * h and h² = ((A - s²) / (2s))² - (s/

2)² . So, V²(s) = (1/9) * s⁴ * h² Substitute h² : V²(s) = (1/9) * s⁴ *
[((A - s²) / (2s))² - (s/2)²]
V²(s) = (1/9) * s⁴ * [(A² - 2As² + s⁴) / (4s²) - s²/4] V²(s) =
(1/9) * s⁴ * [(A² - 2As² + s⁴ - s⁴) / (4s²)] V²(s) = (1/9) * s⁴
* [(A² - 2As²) / (4s²)] V²(s) = (1/9) * s² * (A² - 2As²) / 4
V²(s) = (1/36) * (A²s² - 2As⁴)

Now, substitute A = 5 : V²(s) = (1/36) * (5²s² - 2 * 5 * s⁴) V²(s) =

(1/36) * (25s² - 10s⁴) V²(s) = (25s² - 10s⁴) / 36

2. Differentiate V²(s) with respect to s: Find d(V²)/ds .

d(V²)/ds = d/ds [(25s² - 10s⁴) / 36] d(V²)/ds = (1/36) * d/ds

[25s² - 10s⁴] d(V²)/ds = (1/36) * [25 * 2s - 10 * 4s³] d(V²)/ds
= (1/36) * [50s - 40s³] d(V²)/ds = (10s * (5 - 4s²)) / 36
d(V²)/ds = (5s * (5 - 4s²)) / 18

3. Set the derivative to zero: d(V²)/ds = 0 . Solve this equation for s to find the
critical points.

(5s * (5 - 4s²)) / 18 = 0 This implies either 5s = 0 or 5 - 4s² = 0 . *

5s = 0 => s = 0 . This corresponds to a degenerate pyramid with no volume, so
it is a minimum. * 5 - 4s² = 0 => 4s² = 5 => s² = 5/4 => s = √(5/4) =
√5 / 2 (since s must be positive).
 So, the critical point is s = √5 / 2 .

4. Second Derivative Test: Calculate the second derivative, d²(V²)/ds² . Substitute

the critical s value into the second derivative. If d²(V²)/ds² < 0 , then you have
found a local maximum.

We have d(V²)/ds = (50s - 40s³) / 36 . d²(V²)/ds² = d/ds [(50s -

40s³) / 36] d²(V²)/ds² = (1/36) * d/ds [50s - 40s³]
d²(V²)/ds² = (1/36) * [50 - 40 * 3s²] d²(V²)/ds² = (1/36) * [50
- 120s²]

Now, substitute the critical point s² = 5/4 into the second derivative: d²(V²)/
ds² = (1/36) * [50 - 120 * (5/4)]
d²(V²)/ds² = (1/36) * [50 - 30 * 5] d²(V²)/ds² = (1/36) * [50 -
150] d²(V²)/ds² = (1/36) * [-100]
d²(V²)/ds² = -100 / 36 = -25 / 9

Since d²(V²)/ds² = -25/9 < 0 , the critical point s = √5 / 2 corresponds to

a local maximum volume.

5. Calculate Optimum Dimensions: Once you have the optimal s value, substitute it
back into the equations for h and V to fi## 3. Problem 2: Maximizing Volume with
Fixed Diagonal from Base Corner to Apex

6. Introduction to the Problem: Reiterate the second optimization problem:

maximizing the volume of a square-based pyramid with a fixed length of the
diagonal from a base corner to the apex (e.g., 160cm, 170cm, 180cm).
7. Mathematical Model:
◦ Define variables (e.g., base side length s , vertical height h , volume V ,
diagonal from base corner to apex D ).
◦ Formulate equations for the volume and the relationship between base side
length, vertical height, and the fixed diagonal D .
◦ Express volume as a function of a single variable using the fixed diagonal D
constraint.
8. Optimization Process:
◦ Use calculus (differentiation) to find the critical points of the volume function.
◦ Apply the second derivative test to confirm maximum volume.
◦ Calculate the optimum dimensions (s, h) and the maximum volume for each
given diagonal length.
9. Discussion of Results: Compare the results for different diagonal lengths. Discuss
how this constraint addresses the practicality issue from Problem 1.
Mathematical Details for Problem 2: Maximizing
Volume with Fixed Diagonal from Base Corner to Apex

Variables and Formulas:

Let: * s = side length of the square base * h = vertical height of the pyramid (from the
center of the base to the apex) * V = Volume of the pyramid * D = length of the diagonal
from a base corner to the apex (fixed at 160cm, 170cm, or 180cm)

Formulas: 1. Volume of a square-based pyramid: V = (1/3) * s² * h

1. Relationship between base side length (s) and base diagonal (d): Consider the
square base. The diagonal d divides the square into two right-angled triangles. By
the Pythagorean theorem: s² + s² = d² 2s² = d² s² = d²/2 d = s√2

2. Relationship between h, s, and D: Consider a right-angled triangle formed by the

vertical height h , half of the base diagonal ( d/2 ), and the fixed diagonal D (from
base corner to apex). Half of the base diagonal is (s√2)/2 = s/√2 . By the
Pythagorean theorem: D² = h² + (s/√2)² D² = h² + s²/2

Expressing Volume as a Function of a Single Variable:

We want to maximize V = (1/3) * s² * h subject to the constraint

D² = h² + s²/2 .

From the constraint equation, we can express s² in terms of h and D : s² = 2(D² -

h²)

Substitute this into the volume formula: V(h) = (1/3) * [2(D² - h²)] * h V(h)
= (2/3) * (D²h - h³)

This is the function we need to differentiate with respect to h . Remember that D is a

constant for each case (160cm, 170cm, 180cm).

Optimization Process (using Calculus):

1. Differentiate V(h) with respect to h: Find dV/dh .

We have V(h) = (2/3) * (D²h - h³) . Applying the power rule for
differentiation: dV/dh = d/dh [(2/3) * (D²h - h³)] dV/dh = (2/3) *
[d/dh (D²h) - d/dh (h³)] dV/dh = (2/3) * [D² * 1 - 3h²] dV/dh =
(2/3) * (D² - 3h²)
 2. Set the derivative to zero: dV/dh = 0 . Solve this equation for h to find the
critical points.

(2/3) * (D² - 3h²) = 0 Since 2/3 is not zero, we must have: D² - 3h² =
0 3h² = D² h² = D²/3 h = √(D²/3) h = D / √3 (Since height h must be
positive)

3. Second Derivative Test: Calculate the second derivative, d²V/dh² . Substitute the
critical h value into the second derivative. If d²V/dh² < 0 , then you have found
a local maximum.

We have dV/dh = (2/3) * (D² - 3h²) . Applying the power rule for
differentiation again: d²V/dh² = d/dh [(2/3) * (D² - 3h²)] d²V/dh² =
(2/3) * [d/dh (D²) - d/dh (3h²)] d²V/dh² = (2/3) * [0 - 3 * 2h]
d²V/dh² = (2/3) * [-6h] d²V/dh² = -4h

Since h represents a physical height, h must be positive. Therefore, -4h will

always be negative ( < 0 ). This confirms that the critical point h = D / √3
corresponds to a local maximum volume.

4. Calculate Optimum Dimensions: Once you have the optimal h value ( h = D /

√3 ), substitute it back into the equation for s² (or s ) to find the optimal base
side length.

We have s² = 2(D² - h²) . Substitute h = D / √3 : s² = 2(D² - (D /

√3)²) s² = 2(D² - D²/3) s² = 2((3D² - D²)/3) s² = 2(2D²/3) s² =
4D²/3 s = √(4D²/3) s = 2D / √3 (Since side length s must be positive)

5. Calculate Maximum Volume: Substitute the optimal s and h values into the
volume formula V = (1/3) * s² * h .

V = (1/3) * (4D²/3) * (D/√3) V = 4D³ / (9√3)

Calculations for Different Fixed Diagonal Lengths:

You will need to perform the calculations for h , s , and V using the derived formulas
for each of the given fixed diagonal lengths (D): 160cm, 170cm, and 180cm. Remember
to be consistent with units (convert cm to meters if needed, or work entirely in cm³ and
convert the final volume to m³). This approach provides a clear mathematical method to
find the dimensions that maximize volume for a fixed diagonal from base corner to apex,
which seems a more practical interpretation of your teacher's constraint.
Analysis of Results

Problem 1: Fixed Surface Area (A = 5 m²)

Optimal Dimensions: * Optimal base side length (s): 1.1180 m * Optimal height (h):
1.5811 m * Maximum Volume (V): 0.6588 m³

Discussion: The plot problem1_V_squared_plot.png visually confirms that the

maximum volume (or V²) is achieved at the calculated optimal base side length. The
ratio of optimal height to base side length (h/s) is approximately 1.4142, which is √2 .
This indicates that for a fixed surface area, to maximize the volume of a square-based
pyramid, the height tends to be significantly larger than the base side length. This results
in a relatively tall and narrow pyramid, which might indeed be impractical for a camping
tent as noted by your teacher. While mathematically optimal for volume, such a design
could be unstable, difficult to enter/exit, and offer limited usable internal space for a
person to stand or move comfortably.

Problem 2: Fixed Diagonal from Base Corner to Apex

Optimal Dimensions and Volumes for different D values:

For D = 160 cm (1.6 m): * Optimal height (h): 0.9238 m (92.38 cm) * Optimal base side
length (s): 1.8475 m (184.75 cm) * Maximum Volume (V): 1.0510 m³ (1,051,034.09 cm³)

For D = 170 cm (1.7 m): * Optimal height (h): 0.9815 m (98.15 cm) * Optimal base side
length (s): 1.9630 m (196.30 cm) * Maximum Volume (V): 1.2607 m³ (1,260,676.39 cm³)

For D = 180 cm (1.8 m): * Optimal height (h): 1.0392 m (103.92 cm) * Optimal base side
length (s): 2.0785 m (207.85 cm) * Maximum Volume (V): 1.4965 m³ (1,496,491.90 cm³)

Discussion: The plots problem2_Volume_plot_D_160cm.png ,

problem2_Volume_plot_D_170cm.png , and
problem2_Volume_plot_D_180cm.png visually demonstrate that for each fixed
diagonal D (from base corner to apex), there is a clear maximum volume at the
calculated optimal height. In this problem, the ratio of optimal height to base side length
(h/s) is (D/√3) / (2D/√3) = 1/2 . This means the optimal height is half of the base
side length. This results in a pyramid that is relatively wider and shorter compared to the
solution from Problem 1. This shape is generally more practical for a camping tent,
offering greater stability and more usable internal space, particularly in terms of
headroom near the center and edges. As the fixed diagonal D increases, both the
optimal height and base side length increase proportionally, leading to a larger
maximum volume. This constraint successfully addresses the practicality concern by
ensuring a more reasonable tent shape that can accommodate a person. .85 cm) *
Maximum Volume (V): 1.4965 m³ (1,496,491.90 cm³)

Discussion: The plots problem2_Volume_plot_D_160cm.png ,

problem2_Volume_plot_D_170cm.png , and
problem2_Volume_plot_D_180cm.png visually demonstrate that for each fixed
diagonal D , there is a clear maximum volume at the calculated optimal height. In this
problem, the ratio of optimal height to base side length (h/s) is exactly 0.5. This means
the optimal height is half of the base side length. This results in a pyramid that is
relatively wider and shorter compared to the solution from Problem 1. This shape is
generally more practical for a camping tent, allowing for easier movement and stability.
As the fixed diagonal D increases, both the optimal height and base side length increase
proportionally, leading to a larger maximum volume. This constraint successfully
addresses the practicality concern by ensuring a more reasonable tent shape that can
accommodate a person.

1. Introduction
Optimization is a fundamental concept in mathematics that involves finding the best
possible solution to a problem from a set of available alternatives. This often translates
to maximizing or minimizing a particular quantity, such as profit, cost, area, or volume,
subject to certain constraints. Optimization techniques are widely applied in various
fields, including engineering, economics, computer science, and physics, to solve real-
world problems and improve efficiency.

In the context of design and engineering, optimization plays a crucial role in creating
products that are both functional and resource-efficient. For instance, in the design of
structures, optimization can help determine the dimensions that minimize material
usage while ensuring structural integrity. In manufacturing, it can be used to optimize
production schedules to maximize output and minimize costs. This investigation focuses
on applying optimization principles to a practical design problem: determining the
optimal dimensions of a camping tent.

Camping tents come in various shapes and sizes, each with its own advantages and
disadvantages. The shape of a tent significantly impacts its internal volume, stability,
and material requirements. A common and relatively simple tent shape is the square-
based pyramid. This shape is easy to set up and provides a reasonable amount of
internal space. However, the dimensions of a pyramid tent can vary widely, leading to
different internal volumes even with the same amount of material.
The primary aim of this investigation is to explore how the dimensions of a square-based
pyramid camping tent can be optimized to maximize its internal volume under different
practical constraints. Specifically, the investigation will address two distinct optimization
problems:

1. Maximizing the volume of a square-based pyramid with a fixed surface area,

representing a fixed amount of fabric material.
2. Maximizing the volume of a square-based pyramid with a fixed length of the
diagonal from a base corner to the apex, representing a constraint related to the
usable internal height and space for a person.

By analyzing these two problems, this investigation seeks to understand how different
constraints influence the optimal tent dimensions and to determine which set of
dimensions provides a better balance between maximizing internal volume and
ensuring practical usability for a camping tent. The investigation will utilize
mathematical modeling and calculus techniques, specifically differentiation, to find the
optimal dimensions for each scenario. Geometric formulas for the volume and surface
area of a square-based pyramid will be employed to establish the relationships between
the tent's dimensions and the quantities to be optimized.

The scope of this investigation is limited to the mathematical optimization of an ideal

square-based pyramid shape. It does not account for real-world factors such as the
thickness of the tent material, the space occupied by tent poles, the design of zippers or
vents, or the effects of wind resistance and structural stability beyond the basic
geometric shape. The focus is purely on the mathematical relationship between
dimensions, surface area, and volume under the specified constraints. The findings of
this investigation will provide insights into the theoretical optimal shapes and volumes,
which can serve as a starting point for more complex and realistic tent design
considerations.

4. Comparative Analysis and Conclusion

This investigation explored two distinct approaches to maximizing the internal volume
of a square-based pyramid camping tent: one constrained by a fixed surface area
(representing material) and the other by a fixed diagonal distance from a base corner to
the apex (representing usable internal space for a person). The results from both
optimization problems highlight the significant impact of the chosen constraint on the
optimal dimensions and the resulting tent shape.

In Problem 1, with a fixed surface area of 5 m², the optimization process revealed that
the maximum volume is achieved when the ratio of the tent's height to its base side
length is approximately √2 . This leads to a relatively tall and narrow pyramid. While this
shape is mathematically optimal for maximizing volume given a fixed amount of
material, it poses practical challenges for a camping tent. A tall, narrow tent can be less
stable in windy conditions, and the steep angle of the walls might reduce the usable
floor space and make it difficult for occupants to stand upright comfortably, especially
near the edges of the base. The teacher's comment about the tent being

either "really tall or really flat" likely refers to the outcomes of unconstrained or simply
constrained optimization problems like this one, where the mathematical optimum
might not align with practical design requirements.

In contrast, Problem 2 introduced a constraint based on the fixed diagonal distance from
a base corner to the apex, with values of 160 cm, 170 cm, and 180 cm. This constraint
was chosen to reflect the need for a tent to accommodate a person comfortably. The
optimization results for this problem showed that the maximum volume is achieved
when the height of the pyramid is exactly half of its base side length (h/s = 0.5). This ratio
results in a wider and shorter pyramid shape compared to the optimal shape in Problem
1. This shape is generally more practical for a camping tent, offering greater stability and
more usable internal space, particularly in terms of headroom near the center and
edges. The varying values of the fixed diagonal demonstrate how increasing the required
internal space (represented by the diagonal) leads to larger optimal dimensions and,
consequently, a greater maximum volume.

Comparing the two problems, it is evident that the constraint significantly influences the
optimal tent design. The fixed surface area constraint, while maximizing volume for a
given amount of material, yields a shape that is less practical for a camping tent. The
fixed diagonal constraint, on the other hand, leads to a more practical shape that
balances volume maximization with the need to accommodate occupants. Therefore,
when designing a practical camping tent, simply maximizing volume with a fixed
amount of material is not sufficient; practical considerations, such as usable internal
space, must also be incorporated into the optimization problem.

The research question, "What are the optimum dimensions of a square-based pyramid
camping tent to maximize internal volume with a fixed amount of material?", was
addressed by Problem 1. However, the analysis revealed that the mathematically
optimal solution under this constraint results in an impractical tent shape. By
introducing the second problem with a fixed diagonal constraint, the investigation
explored a more practical approach to tent design optimization. The results from
Problem 2 provide optimal dimensions for tents that are better suited for actual use,
demonstrating that the 'optimum' dimensions depend heavily on the specific criteria
and practical limitations considered.

Limitations and Assumptions:

Several limitations and assumptions were made in this investigation:

• Ideal Geometric Shape: The models assume a perfect square-based pyramid

shape, neglecting the complexities of real tent structures, such as sloping walls
that are not perfectly flat triangles, the presence of tent poles, guy lines, and
stakes.
• Material Properties: The models do not account for the properties of the tent
fabric, such as its elasticity, weight, or how it is cut and joined, which can affect the
actual usable surface area and shape.
• Usable Volume: The optimization maximizes the total geometric volume of the
pyramid. In reality, the usable volume of a tent is less than the total volume due to
the slope of the walls and the space taken up by occupants and gear.
• Fixed Constraints: The constraints (fixed surface area and fixed diagonal) were
treated as exact values. In a real manufacturing scenario, there would be tolerances
and variations in material and dimensions.
• Environmental Factors: The models do not consider external factors such as wind
load, snow load, or terrain, which are critical for tent stability and design.

Extension/Further Research (Optional):

Possible extensions to this investigation could include:

• Analyzing other tent shapes (e.g., dome tents, tunnel tents) and comparing their
volume optimization under similar constraints.
• Incorporating additional constraints related to tent stability, wind resistance, or
ease of manufacturing.
• Developing a more sophisticated model that considers the usable internal volume
rather than the total geometric volume.
• Investigating the effect of material properties and tent component design on the
overall optimization.
• Using computational tools or simulations to explore more complex tent designs
and optimization criteria.

Conclusion:

In conclusion, this investigation successfully applied mathematical optimization

techniques to determine the optimal dimensions of a square-based pyramid camping
tent under two different constraints. The results demonstrate that maximizing volume
with a fixed surface area leads to a tall and narrow, potentially impractical, tent shape.
Conversely, maximizing volume with a fixed diagonal distance from the base corner to
the apex results in a wider and shorter, more practical tent shape. This highlights the
importance of considering practical design constraints in addition to purely
mathematical optimization goals. The findings provide valuable insights into the trade-
offs involved in tent design and offer a mathematical framework for determining
dimensions that balance volume and usability. The investigation also identifies several
limitations and potential areas for further research to develop more comprehensive and
realistic tent optimization models.

5. Bibliography
• [1] (Include any textbooks, websites, or other sources you used here. Remember to
format them consistently and include clickable URLs where applicable.)

6. Appendix (Optional)
• (Include any raw data, detailed calculations, or additional graphs here. You can
refer to the generated plots in this section.)