S00256359 - Mathematical Thinking Project: Optimizing Box Dimensions

This project explores the optimization of box dimensions to minimize production costs, applying mathematical thinking and calculus. The project begins by sketching a box and its net, then derives formulas for volume and surface area, setting the volume to 0.5 cubic meters. It calculates the cost of cardboard based on surface area, using a cost of $0.33 per square meter, and expresses the cost as a function of the box's width. A graph of the cost equation is generated using Desmos, identifying the minimum cost point. Recommendations are made for the box's dimensions, aiming to minimize cost, and these recommendations are validated through calculus by optimizing the cost function. The project concludes by emphasizing the importance of modeling and cost analysis in business decision-making, advocating for companies to model their production processes to reduce costs.

Introduction to Mathematics thinking-project 1
Australian Catholic University
Introduction to Mathematical thinking
Investigative Project
Name:
Student Number:
Course Code: S00256359

Introduction to Mathematics thinking-project 2
(a). The sketch of the box and its net from which it can be constructed are presented below.
(i)The sketch of the box
h
x
(ii)
x
h
In the net above, I have assumed that the box will be closed on top. Have also assumed that the
sides of the square base are x meters each and the height is h meters.
(b) The formula for the volume and surface are of the box. The volume of the box should be the
same as that of the nuts and bolts which will be packed inside it. We will express the formula of
the box as V. The box is a cuboid with square base of dimensions x meters and a height of h
meters cm. Since we have been provided with the volume as 0.5 cubic meters, we will apply the
formula for calculating the volume of a box.
Let the volume= V
The base square sides=x Meters
The height=h Meters

Introduction to Mathematics thinking-project 3
V= l ×w × h
V=( x × x × h¿=x2 h
V= x2 h=0.5 m3
The formula for the surface area of the box is.( There is the assumption that the box will have a
top)
SA= 2 {( l× w )+ ( l× h ) + ( h ×w ) }
SA=2 {( x × x )+ ( x ×h )+ ( h× x ) }
SA=2 x2+4 xh
(c) We then proceed and find the cost of the material used in the making of the box. The card
board used in making the box costs $0.33 per square meter, The cost of making the box will be
obtained by multiplying the surface area of the cost per square meter of card board. Let the cost
be C.
Cost=$0.33(2 x2+4 xh)
But since we know the volume of the box, we simplify and express the height in terms of width
(x meters)
x2 h=0.5m3, h= 0.5
x2
Replacing the height in the cost equation
C=0.33(2x2+4 x( 0.5
x2 )¿
C=0.66x2+ 0.66
x
(d) Drawing the graph of the cost equation using the Desmos graphing calculator.
Using the equation y=0.66x2+ 0.66
x
Thee graph will be as shown below.

Introduction to Mathematics thinking-project 4
(e) Using the above graph of the cost equation, we find that the minimum point is (0.794, 1.247).
Since x is the width of the box and y is the cost incurred in making the box, the minimum cost
incurred in making the box is thus $1.247 while the minimum width of the box is 0.794meters.
Since we have the minimum height
h= 0.5
x2 , h= 0.5
0.7942
h=0.793 meter

Introduction to Mathematics thinking-project 5
(f) The recommendations for the shape of the dimension of the box
The aim of any company is to minimize the cost of production and maximize returns. The
company’s minimum cost of making the box will be considered in this case. A t the minimum
cost of making the box ($1.247), the width and the length of the box is approximately 0.794
meters. This means the box dimensions that can be used to make a box at the lowest cost
possible are for a cube.
To affirm the above results without the use of graphical method, one can use optimization of the
cost function. The main aim in the making of the box is to be economical. Being economical
means that we should make a box that minimizes the cost. To do this, we will have to minimize
the cost function by differentiating it and equating it to zero.
C=0.66x2+ 0.66
x
dC
dx =1.32x- 0.66
x2
1.32x- 0.66
x2 =0
Calculating the value of x that minimizes the cost
x =0.5
1
3 =0.794
h=0.794
We replace x in the cost equation and we obtain C=0.66(0.7942 ¿+ 0.66
0.794
C=$1.247
These results are the same as those obtained from the graph. One can solve the problem by
applying the calculus approach or by use of the graph.
For one to verify the results of the above, we can pick the various points above the obtained
value of x and below the obtained value of x. If we pick 0.8 which is above the obtained value of
x=0.794 meters, we will have C=0.66(0.8*0.9) + (0.66/0.9)=$1.2679 which is higher than the
minimum cost.
As mentioned earlier, a company will operate to reduce the cost incurred in the production of its
product. It thus becomes necessary for any organization to model its products and service
delivery structures. For a company to decide on the best way to handle its production costs, it
needs to look in to various aspects in its production. The modeling above is an example of how

Introduction to Mathematics thinking-project 6
to make a box whose structure is less costly in terms of the materials required. A company needs
to stage various models and make insights for proper decision making. Good models capture the
cost drivers but not elements.

Introduction to Mathematics thinking-project 7