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.

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.

Differentiation

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:

However,

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.

Differentiation

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

Furthermore,

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?

B

A

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.

Mathematical

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?

Nevertheless,

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!!

Beach

Gate

Y

X

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

X

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

X

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

must be positive;

X

X > 0

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

To find the maximum value of X :

X

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

= 2 – 2,950,000X-2

Solving the above equation, we get the values of X

as:

2 – 2,950,000X-2 = 0

2 = 2,950,000X-2

2X2 = 2,950,000

X2 = 2,950,000 / 2

X

= 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:

X

T(x) = 2X +

2950000

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

X

? C(X) = 4.84 ( 2X + 2950000

)

= 9.68X +

14278000/X

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

C(X)

X

0

Minimum

point

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

building.

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 +

30

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

X-intercepts.

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

wide.

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

distance)

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)

30

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