Mathematics able to understand the concept of optimization but


Mathematics is a very essential part of
our daily routine as we have mathematics everywhere around us; like for
measuring ingredients while cooking, counting money, measuring fuel used by our
cars etc. However, mathematics is also very important in our studies regardless
of the subject as every subject has its own use of mathematics. In my experience,
we use mathematics in science and technology the most as it eases our work and
makes the procedure of an experiment or research simpler.

During my IB course, ‘Information
Technology in Global society’ was one of my subjects in which we had to make a
product using official softwares according to our client’s request, for our
internal assessments. I had chosen AutoCAD architectural 2014 to make my
product and I planned to make an architectural plot design of a resort
considering my client’s requirements. I thought it would be easy to just draw
the design like I had drawn on a paper but I was wrong. We had to use exact
measurements of the buildings and arcs of the roofs to draw the structures. However,
I did not know how to make optimal use of the land and make it economical for
my client.

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Hence, I decided to use optimization and
implicit differentiation in Calculus to help me make a standard plan of the
plot. In addition, using optimization I shall be able to utilize the land more
efficiently with additional structures like water features and landscaping. Optimization
enables us to find the maximum and minimum part of a model or a given function
that helps us to decide how the real structure will look like.

My inspiration to take up this topic was
created in our mathematics class when I was able to understand the concept of
optimization but I was finding difficulties analyzing its applications and
solving problems thus, I decided to research this topic further and use it for
my Mathematics’ internal assessment. Furthermore, my main aim of exploring this
topic is to demonstrate how calculus can help solve technical problems – especially
engineering. However, calculus is also widely used in other faculties including
religious and indigenous knowledge systems but I shall only consider calculus
used in the architecture and civil engineering world to explore my topic.

















What is Calculus and what are its practical uses?

Firstly, to start evaluating the
research question further, it is required to know – what is calculus? We know
that calculus is a mathematical method and a name of a mathematician who
invented this method to solve mathematical issues related to rate and accretion
with more ease. Therefore, scientists define calculus as a study of rate of
change and accumulation.

Calculus is not a new concept in
mathematics like geometry and algebra but an extended combination of both of
these by using the idea of limits. Limits allow us
to study what happens when points on a graph get closer and closer together
until their distance is infinitesimally small (almost zero). Once the idea of
limits is applied to the Calculus problem, the techniques used in algebra and geometry can
be implemented.







is given in f(x) function form or in an equation form, where x is a real
variable. To find gradient of a curve we use the same formula that we use in
geometry to find gradient of a linear graph:

this formula is substituted to find the gradient of the curve with the function
y=f(x). Finding gradient this way is called derivation, when we derive a
function we find the gradient of its graph at the given point. Geometrically, the derivative of f at the point x = a is the
slope of the tangent line to the function f at
the point a. Which means the gradient of the tangent line and the curve at this
point is same. An example of derivation; f(x) = x2 so derivative/
gradient of this curve is f'(x) = 2x.

Calculus can be used everywhere in our daily life; while cooking, solving
economical issues, calculating amount of fuel used by car etc. Moreover, it is
also used in engineering and medical world and has been found to be very
beneficial so far. In medical world, calculus can be used to find the
exponential growth of a certain disease or to find how a patient can be treated
with minimum harm. Similarly, in engineering, calculus can be used to create
mathematical models in order to arrive into an optimal solution.

Examples of some
engineering and exponential growth graphs


considering calculus’ vast benefits the architectural world is also making good
use of it. Differentiation can be used to find maximum space available and how
can it optimally be used. By using optimization in differentiation, we can
easily find the maximum area of a given plot and its dimensions thus design the
structure accordingly.











What is optimization in calculus and how can it be used in
architectural world?



Optimization in mathematics means the selection of a
best element from some set of available alternatives. In the science world, it
is also referred as mathematical programming since it consists of maximizing or
minimizing an actual function. This is done by systematically choosing domain
values from within an allowed set. For example, if a farmer wants to know the
optimum space that he can attain from fencing three sides around a wall by a 3m
fence for his sheep. Then the easiest way of doing this is using optimization.



Optimization is a branch of applied mathematics, which is useful in many
different fields. Some examples include; Manufacturing, Production, Inventory
control, Transportation, Scheduling, Networks, Finance, Engineering, Mechanics,
Economic, Control engineering, Marketing and Policy Modeling. The most common use
of optimization is while designing a model before engineering it. Applying
optimization to models can give an idea, to the manufacturing engineers, how
will the actual product look like. Some civil engineering problems that are
solved by optimization include cutting and filling of roads, life-cycle
analysis of structures and infrastructures, resource leveling and schedule
optimization – this is also applicable for architecture as architects use
optimization while designing the plot or other structures.

The basic optimization problem consists
of: the objective function, f(x), which is the output we try to maximize or
minimize. Variables, x1 x2 x3 and so on, which
are the inputs – things we can control. They are abbreviated xn to
refer to individuals or x to refer to them as a group. Constraints, which are
equations that place limits on how big or small some variables can get.
Equality constraints are usually noted hn (x) and inequality constraints are
noted gn (x).











How can we use optimization in differentiation calculus to design an
optimal architecture of a plot?

calculus has many applications but my main aim is to find how I can use
implicit differentiation and optimization to design an optimal architectural
plot for my dream resort. While drawing the plot, I realized that I am
occupying a lot of space by some small structures in some places whereas
leaving out a lot of land for landscaping while it can be used for extra
recreational activities. The perfect use of land is very crucial; economically and
environmentally. Therefore, I thought of using optimization to re-plan my
structures on my design to make good use of land.

Design of my optimized plot in
AutoCAD 2014













Optimization used to draw such designs
is called design optimization. Design optimization is the process of finding
the best design parameters that satisfy project requirements.  Engineers typically use design of experiments
(DOE), statistics, and optimization techniques to evaluate trade-offs and
determine the best design. Design optimization often involves working in multiple
design environments in order to evaluate the effects that design parameters
have across interrelated physical domains.

Similarly, I want to
use the excelled optimization methods to helve me design my plot. Since, in the
procedure of constructing we need to do the fencing of the area, we purchased,
first so I thought I should find the cost and of course the area required to be
fenced first also. Therefore, first I drew the coast were our land belonged and
measured our part from it. Here are few pictures of the coast and our land:
















However, this is not all our land, our part is only 1.0
acres x 1.0acres. Hence, to know this part of land I used the basic method of
optimisation to obtain it.  The polygon
in the picture below highlights our part of land on the coast.









I started by finding the area of the
part that will be used for the construction:

The dimensions of the entire plot were such: 1,970.95m
x 1,497.00m

Hence, the area of the plot is 2,950,512.15m2   ? 2,950,000.00m2

Therefore, I need enough fence to cover this
area thus I researched the cost of fencing 10m so when I find the optimal dimensions to cover the above area of
land I can work out the total cost of overall fencing. Furthermore, I found out
that the price for a square meter is $9.84 hence I need to buy:

1m2  – $4.84      ?     
2,950,000.00m2 = $14,278,000.00

However, this is the
maximum value that can possibly be but it is not the real cost because one side
will be the beach thus that side would not require fencing and the land will be
further optimized according to the optimization laws so the total area shall
also vary which will change this cost  –
anyways it looks too expensive!!















The figure above shows the outline of
the plot that requires to be optimized.  However,
not all sides need fencing so I need to subtract that part to calculate the
exact length that requires fencing. So I assumed the width to be X and length
to be Y, hence the perimeter (P) and area (A) would be:

Perimeter, 2X + Y = P

Area, XY = A and A = 2,950,000 m2

So XY = 2,950,000


Making Y the subject we get; Y = 2,950,000m2



Therefore the equation will be P = 2X + 2,950,000m2                 and since the lengths
must be positive;


X > 0      
and          2X + 2,950,000m2       > 0



To find the maximum value of X :


dY/dX = 2X + 2,950,000m2

= 2 – 2,950,000X-2


Solving the above equation, we get the values of X


2 – 2,950,000X-2 = 0

2 = 2,950,000X-2

2X2 =  2,950,000

X2 = 2,950,000 / 2

=       1,475,000

Therefore X = 1214.5m which is the
maximum value of X.

We can prove this by using the second
derivative test and the sign diagram test:

dY/dX  = 2 – 2,950,000X-2

d2Y/dX2 = –

And by sign diagram:






Therefore, above we found the total perimeter,
which was total length:


 T(x) = 2X +


Hence, from here we need to find the
cost function, C(x). We know that for 1m of fencing, the cost is $4.84 and our total
perimeter is P so,

1m – 4.84  so Pm – 4.84P


?  C(X) = 4.84 (  2X + 2950000 

            = 9.68X +

C'(X) = 9.68 -14278000X-2

From this function we can determine the minimum cost by
plotting a graph of C'(X) versus X as we already know the value of X:

X = 1214.5













Hence, from the sigmoid curve above we know that the minimum
value is:

Thus the minimum cost is -14278000X-2 + 9.68 when
X =

Which is

After using optimization for fencing and its cost, I
considered using it for guest building structure and roofing. Optimization
gives exact and best size hence I used the following method for modeling the


The different elevations of the building and the roof design


The building above has a rectangular base with 1990m length
and 867m width, with the maximum height 30m to 50m. The roof structure is
modeled using parabola.

Assuming one side of the building to be on axis when maximum
height of the building is taken to be 30m and keeping the origin at the
midpoint of the shorter side of the base – the parabolic model of the roof
structure has vertex on the point (0,30) and the other end points on (-30,0)
and (30,0).

Therefore, the model for roof when the building has height of
30m forms the following equation:

Y = a(X – 0)2 + 30

   = aX2 +

By using the values of endpoints, we get the parabola’s

Y = aX2 + 30

   = a (30)2
+ 30

  = 900a + 30

  So, a = -1/30

Hence, when building has the height of 30m the roof structure
is modeled by the following equation:

Y = (-1/30)X2 + 30

The equation gives the following graph:

As we can see the parabola is symmetrical with respect to Y
and has (0,0) as its midpoint. The graph shows the height of the building.

Furthermore, I found the dimensions of the building with the
maximum volume, which shall fit under this roof. I did this by using the
equation above and the known dimensions of the building – 1990m long and 867m

V = 2X x 1990 x Y

   = 2X x 1990 x
((-1/30)X2 + 30)

   = ((-398/3)X2
+ 119400X)

To find the maximum volume we differentiate by deriving the
volume equation:

V’ = ((-398/3)X3 + 119400X)

    = -398X2
+ 119400

And when equating it to zero we get:

V’ = 0

0 = -398X2 + 119400

398X2 = 119400

X2 = 119400/398

X =       300   

   = 17.3m ( as X is

Thus the value of Y is:

Y = ((-1/30)X2 + 30

   = ((-1/30)(17.3)2
+ 30

   = 20

As resulted above, the origin is on the midpoint of the base,
the width of the building with maximum volume fitting under the roof with
parabola of height 30m is

2X = width

2 (17.3) = 34.6m

Hence the dimensions of the building with the parabolic roof
are: 34.6 x 20 x 1990 (all in meters).

Therefore, the façade of the building is such:

However, the actual height is ranged between 30m to 50m so we
need to find how changing the height of the roof affects the building’s dimensions.

So we let H = total height of the roof so the vertex on
parabola becomes (0,H) giving the equation:

Y = bX2 + H  {X2 from before}

Therefore at the x-intercept (30,0) the equation becomes:

0 = b(30)2










How changing height affects the dimensions of the building:

Total height
of the roof (H)

Total height
of building(B)