PROJECT TITLE:
Using differential calculus to determine
the maximum area of a court for a
sports club, using 1000 metres of
fencing.
1
TABLE OF CONTENTS
TITLE ........................................................................................................................................ 3
PROBLEM STATEMENT ....................................................................................................... 3
MATHEMATICAL FORMULATION..................................................................................... 4
Diagrams: .....................................................................................................................................4
Variables: .....................................................................................................................................4
(i)
CALCULUS METHOD ........................................................................................................6
(ii)
GRAPHICAL METHOD .................................................................................................7
PROBLEM SOLUTION ........................................................................................................... 8
Assumptions: ................................................................................................................................8
CALCULUS CALCULATIONS: ..................................................................................................8
GRAPHS .................................................................................................................................... 12
APPLICATION OF SOLUTION ........................................................................................... 16
DISCUSSION OF FINDINGS .............................................................................................. 17
CONCLUSION........................................................................................................................ 18
SUGGESTIONS FOR FUTURE ANALYSIS ....................................................................... 19
REFERENCES ....................................................................................................................... 20
2
TITLE
Using differential calculus to determine the maximum area of a court for a sports club, using
1000 metres of fencing.
PROBLEM STATEMENT
In recent times, it was found that there had been a substantial rise in gang violence and crime
amongst youths in communities. To curb this problem the leaders of the community have
introduced many sporting activities which target the youths with hopes to offer them various
opportunities to better themselves. With this initiative the community has decided to erect a
sports club, however, there has been an ongoing dispute on which design, A or B, would offer
more area inside the sports club’s court. The purpose of this project is to determine which design
for a court would offer maximum area, utilising 1000 metres of fencing for each design.
3
MATHEMATICAL FORMULATION
Diagrams:
•
Diagram Showing Court Design A
3x
2y
•
Diagram Showing Court Design B
x
x
y
y
Variables:
•
Let ‘L’ be the length of the courts.
•
Let ‘W’ be the width of the courts.
•
Let ‘A’ be the general area of the courts.
4
x
•
Let ‘P’ be the general Perimeters of the courts.
•
Let ‘AA’ be the area of court design A.
•
Let ‘AB’ be the area of court design A.
•
Let ‘𝑑𝑥 ’be the first differential and gradient function for each equation of Area with respect
𝑑𝐴
to 𝑥.
•
𝑑2 𝐴
Let ‘𝑑𝑥 2 ’ be the second differential for each equation of Area with respect to 𝑥.
5
CALCULUS METHOD
(i)
To obtain the design which gives the maximum area of a court when 1000m of fencing is used to
enclose it:
•
Firstly, an equation for area of both designs will found by using ‘A= LxW.’
•
An equation for both perimeters, P, will also be calculated and equated to 1000m. ‘y’ will be
made the subject of the formula, and will be substituted into the equation of area. Hence, two
equations written with A in terms of 𝑥 will be obtained.
𝑑𝐴
•
The areas will then be differentiated to obtain the first differential,
•
For stationary values [maximum or minimum], the first differential, 𝑑𝑥 , will be equated to
𝑑𝑥
.
𝑑𝐴
zero [0] to find the values of 𝑥 since the turning/stationary points are the maximum or
minimum values.
•
To determine whether the values obtained are maximum or minimum values, the second
differential,
value of
𝑑2 𝐴
𝑑𝑥 2
, will then be found. When 𝑥 is substituted into the equation, a negative
𝑑2 𝐴
𝑑2 𝐴
will indicate that the area is a maximum value. A positive value of 𝑑𝑥 2 will
𝑑𝑥 2
indicate that the area is a minimum value.
6
•
The value of 𝑥 for each area equation, which gives a negative value of
𝑑2 𝐴
𝑑𝑥 2
, will then be
substituted to calculate the maximum areas.
•
The two final values of maximum area, AA and AB , will then be compared, and the court
design with the higher value will be chosen.
GRAPHICAL METHOD
(ii)
To obtain the design which gives the maximum area of a court when 1000m of fencing is used to
enclose it:
•
Firstly, an equation for area of both designs will found by using ‘A= LxW’.
•
An equation for both perimeters, P, will also be calculated and equated to 1000m. ‘y’ will be
made the subject of the formula, and will be substituted into the equation of area. Hence, two
quadratic equations, of the form A= 𝑎𝑥 2 + 𝑏𝑐 + 𝑐, will be obtained.
•
The axis of symmetry for each equation of area will then be calculated by using
𝑥 = −𝑏/2𝑎.
•
The x-axis intercepts will be found by equating ‘A’ in each area equation to zero [0].
7
•
A table of values will be created using 20 pairs of values of A and of 𝑥, inclusive of each axis
of symmetry, and the two x-axis intercepts for each equation. The two graphs will then be
plotted.
•
The two graphs will be compared, and the court design with the higher maximum point value
(of A) will be chosen.
PROBLEM SOLUTION
Assumptions:
1) All of the 1000 metres of fencing was used in each design.
2) The gradient of the terrain for each court design was 0 throughout (since they were flat
surfaces), hence the area and perimeter for a standard rectangular shape could be calculated.
CALCULUS CALCULATIONS:
Design A
AA = L × W
P = (L + B) × 2
AA = 3x × 2y
P = (3x + 2y) × 2
AA = 6xy ……….(1)
P = 6x + 4y
Recall: 1000m of fencing is used for each design
∴ P = 1000m
∴ 6x + 4y = 1000m
8
4y = 1000m – 6x
(divide above equation by 4)
y=
1000m−6x
4
(simplify)
y=
500𝑚−3𝑥
2
……………. (2)
Substituting equation (2) into equation (1)
∴ AA = 6x
(500𝑚 − 3𝑥)
2
3000x − 18x 2
2
AA =
AA = 1500x − 9x 2
First Differential,
dAA
dx
= (1500)(1 × x1−1 ) − (9)(2 × x 2−1 )
dAA
= 1500 − 18x
dx
At stationary point,
dAA
dx
=0
∴ 1500m − 18x = 0
-18x = -1500m
(divide throughout by -18)
1
x = 833m
To ensure that the area is maximum, we find
∴
d2 AA
dx2
d2 AA
dx2
= −18(1 × 𝑥1−1 )
9
∴
d2 AA
dx2
= −18
As illustrated,
d2 AA
dx2
< 0 (-18 < 0), hence the area 𝐴𝐴 is maximum
1
To find the corresponding value of y, substitute x = 833m into equation (2)
∴y=
1
3
500m−3(83 m)
2
y = 125 m (by calculator)
1
Hence, AA = 6(83 3 m)(125m) from equation (1)
AA = 62500 m2
Therefore, the maximum area of the court Design A is 62500 m2
Design B
A=𝐿 ×𝑊
P = 3L + 4W
A = (y + y) × [𝑥 + 𝑥 + 𝑥]
P = 3[x + x + x] + 4[y + y]
A = 2y × 3𝑥
P = 3[3x] + 4[2y]
A = 6xy……….. [1]
P = 9x + 8y
Since 1000m of fencing is used
P = 1000m
9x + 8y = 1000m
8y = 1000m – 9x
(÷ 8)
y=
1000𝑚 −9𝑥
8
………… [2]
Substituting equation [2] into equation [1]
10
1000−9𝑥
AB = 6x (
8
)
6000𝑥 −54𝑥 2
AB =
8
3
∴ AB = 750x - 6 4 𝑥 2
First differential
∴
𝑑𝐴𝐵
𝑑𝑥
𝑑𝐴𝐵
𝑑𝑥
3
= 750(1 × 𝑥1−1 ) − 6 4 (2 × 𝑥 2−1 )
1
= 750𝑚 − 13 2 𝑥
𝑑𝐴
At a stationary value, 𝑑𝑥 = 0
1
750m - 13 2 𝑥 = 0
1
−13 2 𝑥 = -750m
1
(divide above equation by -13 2 𝑥)
5
∴ x = 559m
To ensure that the area is maximum, we find
𝑑2 𝐴𝐵
𝑑𝑥 2
𝑑2 𝐴𝐵
𝑑𝑥 2
𝑑2 𝐴𝐵
𝑑𝑥 2
1
= 0 - 13 2 (1× 𝑥1−1 )
1
= - 13 2
As illustrated
𝑑2 𝐴𝐵
𝑑𝑥 2
1
< 0 ( i.e. - 13 2 < 0) hence the area is a maximum.
5
To find the corresponding value of y, substitute x = 559m into equation of (2)
y=
5
9
[600 −9(55 ) ] 𝑚
8
1
y = 622 m (by calculator)
5
1
Hence A = 6(559m) (622 m)
11
1
AB = 208333 𝑚2
1
Therefore, the maximum area of the court design B is 208333 𝑚2
GRAPHS
DESIGN A
Graph of AA = 1500x − 9x 2
•
𝑏
Axis of Symmetry Calculation: 𝑥 = − 2𝑎
𝑥=
−1500
−18
1
𝑥 = 833m
•
Table of Values sample calculation: A = 1500x − 9x 2
12
1
1
1
When 𝑥 = 83 3 m , A = 1500 ( 83 3 m ) − 9( 83 3 m )2
A= 62500 m2
1
1
Therefore, the co-ordinate is (833m, 62500), which is of the form (𝑥, 𝐴), 𝑥 = 83 3 m &
𝐴 = 62500.
Table of Values (correct to 2 decimal places) for Design A (AA = 1500x − 9x 2 ):
𝑥
A
0.00
0.00
10.00
14100.00
20.00
26400.00
30.00
36900.00
40.00
45600.00
50.00
52500.00
60.00
57600.00
70.00
60900.00
83.33
62500.00
100.00
60000.00
110.00
56100.00
120.00
50400.00
130.00
42900.00
140.00
33600.00
150.00
22500.00
160.00
9600.00
166.66
0.00
13
DESIGN B
3
Graph of AB = 750x - 6 4 𝑥 2
•
𝑏
Axis of Symmetry Calculation: 𝑥 = − 2𝑎
𝑥=
−750
3
2 × 64
𝑥 = 55
•
5
9
3
Table of Values sample calculation: A = 750x - 6 4 𝑥 2
5
5
3
5 2
When 𝑥 = 55 9, A = 750(55 9) – 6 4 (55 9)
1
𝐴 = 208333
14
5
1
5
Therefore the co-ordinate is (55 9, 208333), which is of the form (𝑥, 𝐴), where 𝑥 = 55 9 &
1
𝐴 =208333
.
3
Table of values (correct to 2 decimal places) for Design B (AB = 750x - 6 4 𝑥 2 ):
𝑥
A
0.00
0.00
5.00
3581.25
15.00
9731.25
25.00
14531.25
35.00
17981.25
45.00
20081.25
55.00
20831.25
55.55
20833.33
65.00
20231.25
75.00
18281.25
85.00
14981.25
95.00
10331.25
105.00
4331.25
111.11
0.00
15
APPLICATION OF SOLUTION
Differentiation was used to determine the gradient function of the curve. At the stationary point
the gradient function,
𝑑2 𝑦
𝑑𝑦
, was equated to zero. The second differential, 𝑑𝑥 2 , was then used to
𝑑𝑥
determine the nature of the stationary points. If the second differential was positive, then the
value of x would be at minimum and if negative the value would be at a maximum. It was found
that the second differential was -18 for design A which meant that the value of x was at a
maximum. Hence, the maximum area of Design A was found to be 62500 𝑚2 . Design B had a
1
second differential of -13 2 which meant that the value of x was at a maximum. Hence, the
1
maximum area of Design B was found to be 208333 𝑚2
The graph was also used to determine the maximum point. From the graph Design A was found
1
to have a maximum area of 62500 𝑚2 with an axis of symmetry 𝑥 = 83 3. Design B however
1
5
was found to have a maximum area of 208333with an axis of symmetry of 𝑥 = 559. These are
the same values obtained from the previous calculations which confirms the calculus method.
16
DISCUSSION OF FINDINGS
The intent of this project was accomplished via both the calculus and graphical methods.
The calculus method was used to determine the maximum area of the court that could be fenced
off by 1000 metres (m) of fencing. Two standard equations were obtained, each of perimeter and
of area, by use of the diagrams shown in mathematical formulation. The perimeter, P, was
equated to 1000m, and stated with y as the subject, since the fencing used in each design was
1000m. For design A, the determined equation of perimeter was y =
the equation was 𝑦 =
1000−9𝑥
8
500−3𝑥
2
whereas in design B
. These equations were substituted into the calculated
equation of area, A = 6xy. Hence, by solving simultaneously, two quadratic equations, AA =
3
1500x − 9x 2 , and AB = 750x - 6 4 𝑥 2 , were obtained for both areas. The two quadratic equations
were then differentiated with respect to A, and equated to 0, since at a stationary point, the
𝑑𝐴
gradient is always 0 (𝑑𝑥 = 0). Therefore,
dAA
dx
was found to be 1500 − 18x,
1
and was equated to 0. In this case, the dimension ‘x’ was calculated to be 833 m by simple
algebra. Similarly,
𝑑𝐴𝐵
𝑑𝑥
1
was found to be 750 − 13 𝑥, and equated to 0. The dimension of ‘x’ in
2
5
this instance was 559 m. After, the second differentials were to be determined, and if negative in
value, a maximum area and value of x would have been confirmed. In both cases, the second
differentials,
d2 AA
dx2
and
𝑑2 𝐴𝐵
𝑑𝑥 2
respectively, both generated negative values, thus confirming that
the dimension x obtained was a maximum for the area. Then the separate dimensions, y, for each
design were found by substituting the respective values of x into the equations of perimeter,
written with y as the subject. These values of y for Design A and Design B were calculated to be
17
1
‘y’ = 125 m and ‘y’ = 622 m respectively. Lastly, the two areas, AA and AB were calculated by
substituting the corresponding values of x and of y, and the results were compared.
The graphical method was an alternative method used to determine and verify that the maximum
values of x, of y, and of area obtained in the calculus method were true. Two graphs for both
quadratic equations found (for AA and AB) were plotted, and the maximum areas were read off
accordingly.
CONCLUSION
By using the differential calculus method, and graphical method, the maximum areas of the
sports club’s court designs A and B, using 1000 metres of fencing each, were determined to be
1
1
62500 m2 and 208333 𝑚2 respectively (i.e. AA=62500 m2 AB = 208333 𝑚2). For Design A, the
1
maximum values of the dimensions were found to be 𝑥 𝑚 = 83 3 𝑚 and y m= 125 m , and for
5
1
Design B, 𝑥 𝑚 = 55 9 𝑚 and y m = 622 m. Therefore, by comparing the results obtained, the
1
area of Design A was greater than the area of Design B (i.e. 62500 m2 > 208333 𝑚2 ). Hence, the
sports club’s court to be constructed should be Design A, since it offers the greater area.
18
SUGGESTIONS FOR FUTURE ANALYSIS
The uses of differential calculus are not only restricted to determining the maximum area and
dimensions of an object. For example, differential calculus can also be used to determine the
minimum or maximum volume of standard objects (given a specific surface area), the minimum
amount of material needed to be used to build an object, or the maximum perimeter of an object
for which the area is a maximum (which could have been calculated in this project).
Furthermore, with respect to this project, the area of a court using 1000 metres of fencing could
have been maximized via the use of a different design, that is, many other court shapes could
have been used instead of the standard rectangular shape.
19
REFERENCES
•
Khan Academy. (n.d.). Retrieved from https://www.khanacademy.org/.
•
PASS Maths: CSEC Math and Add Math Past Papers and Solutions. (n.d.). Retrieved from
https://www.faspassmaths.com/add-math-study-guide.
•
Talbert, J. F., & Heng, H. H. (2008). Additional mathematics: pure & applied. Singapore:
Pearson Longman.
20