International assessment 2

INTERNATIONAL ASSESSMENT 2

IB Mathematics HL Year 2

Rie Yamada

(0,0)(-36,0)

(0,h)(x,y)

(36,0)

150m

(0,h)(x,y)

(0,0) (36,0)(-36,0)

- Modeling a functional building -

The building has a rectangular base 150m long and 72m wide as shown in the diagram 1 below. The

maximum height of the structure should not exceed 75% of its width for stability or be less than half the

width for aesthetic purposes. The minimum height of a room in a public building is 2.5m

Diagram 1: The building with a roof structure

As the height of the structure should be between 50% and 75% of its width, the height can range from 36m

to 54m. The diagram below represents a model for the roof structure when the height is 36m.

Diagram 2: The roof structure with the height of 36m

As the roof structure is identical to a parabola being flipped down, the equation of a parabola with a negative

slope can be applied for the roof structure.

y = ax2 +bx +c

0 = 362a + 36b + h

0 = (-36)2a – 36b +h

y = h

As the calculation shows, the equation for a roof structure is y = −h

1296x2 + h. [equation 1]

0 = 362a + 36b + h0 = 362a – 36b + h0 = 72bb = 0

0 = 362a + 36b + h0 = 362a – 36b + h0 = 2(36)2a + 2h1296a = h

a = −h

1296∴y =

−h1296

x2 + h

(0,0)(-36,0)

(0,h)(x,y)

(36,0)

Z

When the height is 36m, the equation for the roof structure is y = −136

x2 + 36.

Since the structure is a building which contains several cubic rooms inside, a cuboid needs to fit inside the

roof structure with maximum volume in order to utilize the space. The diagram 3 demonstrates a cuboid

with maximum volume.

Diagram 3: The cuboid with maximum volume

The dimensions, the volume and the floor area change since its height can vary from 36m to 54m. However,

the width represented by z does not change because it is kept constant.

Since the height is known, the three dimensions of the cuboid can be calculated using the equation 1 above.

As the volume is at its maximum, the value of x can be calculated by finding when the derivative of the

volume equation is zero. Then the value of y can also be found by plugging x in the equation 1. Also the

volume can be easily calculated by multiplying all the three dimensions of the cuboid. The two sample

calculations using the heights of 36m and 54m are shown below. The rest of the floor areas for different

heights are calculated in the same way and are illustrated in the table 1 on page 5.

A = 2xy

V = 2xy×150

= 300 x(−hx21296

+ h)

= −25h

108x3 + 300xh

= −25h

108x3 + 300xh

The volume of the cuboid differs for each height of the structure and it can be found by the equation above.

Now, by using the equation for the volume, the equation which shows the value of x for different heights can

be found.

V = −25h

108x3 + 300xh

dVdx

=−25h36

x2

+300h

0 = −25h36

x2

+300h

300h = 25h36

x2

300 = 2536x2 x2=432

x=√432

x=12√3

X is a constant value 12√ 3 at all the heights . Thus, the width of the base of the structure does not change

although its height varies.

The equation which shows the value of y for different heights can be also found by using the equation for

the roof structure (equation 1 above) as x stays constant for all heights

y = −h

1296x2 + h

= −432h

1296+h

= −h3

+h

= 23h

As the equation shows, y is always two-third of the height of the structure.

The examples below are the sample calculations for the dimensions and the volume of the cuboid, when the

height is 36m and 54m.

(1) When the height is 36m

The base of the shaded area in the diagram 3 is 2x and the height is y. The equations for the shaded area

and the volume are shown below. This equation can be applied for any height raging from 36m to 54m.

h = 36m V = −50

6x3 + 10800x

dvdx

= - 25x2 + 10800

0 = - 25 x2 + 10800

25 x2 = 10800

x2 = 432

x = 12√ 3

y = −hx21296

+ h

= −(12√3)2

36 + 36

= 24


h = 54m V = −25

2x3 + 16200x

dvdx

= −75

2x2 + 16200

0 = −75

2 x2 + 16200

752

x2 = 16200

x2 = 432

x = 12√ 3

y = −(12√3)2

24 + 54

∴ x = 12√3 y = 24 Volume = 149649.2m3

= 36

∴ x = 12√3 y = 36

Volume = 224473.8m3

V = 300xy= (12√3)×24×300= 86400√3= 149649.2

V = 2xy×150= (12√3)×36×150= 64800√3= 224473.8

Table 1: The dimensions, the volumes and the floor areas for different height

Heigh

t X Y Z Volume Floor Area

36 20.78461 24 150

149649.

2 56118.45

37 20.78461

24.6666

7 150

153806.

1 56118.45

38 20.78461

25.3333

3 150 157963 62353.83

39 20.78461 26 150 162120 62353.83

40 20.78461

26.6666

7 150

166276.

9 62353.83

41 20.78461

27.3333

3 150

170433.

8 62353.83

42 20.78461 28 150

174590.

7 68589.21

43 20.78461

28.6666

7 150

178747.

6 68589.21

44 20.78461

29.3333

3 150

182904.

6 68589.21

45 20.78461 30 150

187061.

5 74824.59

46 20.78461

30.6666

7 150

191218.

4 74824.59

47 20.78461

31.3333

3 150

195375.

3 74824.59

48 20.78461 32 150

199532.

3 74824.59

49 20.78461

32.6666

7 150

203689.

2 81059.98

50 20.78461

33.3333

3 150

207846.

1 81059.98

51 20.78461 34 150 212003 81059.98

52 20.78461 34.6666 150 216159. 81059.98

7 9

53 20.78461

35.3333

3 150

220316.

9 87295.36

54 20.78461 36 150

224473.

8 87295.36

This table shows the dimensions, volumes and floor areas for different height, which are calculated by the

way explained in the sample calculations. The values of x stay constant for all heights, but the values of y

changes by two-third as the height increase by one. As the value of y changes, the volume and the floor area

of the cuboid also change.

Now, investigate how changes to the height of the structure affect the dimensions of the largest possible

cuboid and its volume. Both the table 1 on page 3 and the diagram 4 below show the pattern of the changes

in width and height of the cuboid.

Diagram 4: How changes in height affect width and height of cuboid

The diagram 4 illustrates how changes in height affect the width (x) and the height (y) of the cuboid. No

matter how the height of the roof structure changes, the width of the cuboid is always at 12√3 in order to

have the maximum volume. On the other hand, the height of the cuboid changes as the height of the roof

structure increases. As the table 1 shows, the height of the cuboid is always two-third of the height of the

roof structure. Thus, the height constantly increases as the height of the structure increases by 1m each time.

Similarly, the volume of the cuboid changes as the height of the structure is altered.

Diagram 5: How changes in height affect volume of cuboid

The graph above shows how changes in the height of the roof structure affect the volume of the cuboid. As

the height of the structure increases, the volume of the cuboid also increases. This is because while the width

(x) and the length (z) stay constant, the height (y) increase as the height of the structure increases. As the

equation for volume is V = 300 xy, V changes when a component of it changes. Therefore, the architect can

increase the volume of the building by raising its height.

(0,h)(x,y)

(0,0) (36,0)(-36,0)

2.5

Diagram 4: Floor Areas in the cuboid

In addition, as the minimum height of the room is 2.5m, the building can increase the number of rooms until

the total heights of rooms do not exceed the height of the building. In this case, the rooms are cuboid with

the width of x and the height of y, the building can have rooms until the multiple of 2.5m is about to exceed

the value of y. For example, the height of the cuboid is 36m when the height of the roof structure is 54m. As

the height of each room is 2.5m, 14 rooms can be created (36m÷2.5m =14.4 rooms). The total floor area in

the cuboid is the sum of an area for each floor. In this case, the area for each floor is constant as the rooms

are inside a single cuboid. Thus the floor area is calculated by multiplying an area for one floor by the

number of floors which can fit within the roof structure. The two sample calculations for the heights of 36m

and 54m are shown below.

Room number = height of cuboid ÷ 2.5 (rejecting the decimal places)

Total floor area = (2x × 150) × room number


24m÷2.5m =9.6 The cuboid can have 14 rooms at maximum.

As the area for each floor is:

A = 2x×150

= 300 (12√3)

= 360√3

The total floor area is:

TA = 9 (360√3)

= 3340√3

= 56118.45

∴ When the height is 36m, Floor Area = 56118.45


36m÷2.5m =14.4 The cuboid can have 14 rooms at maximum.

As the area for each floor is:

A = 2x×150

= 300 (12√3)

= 360√3

The total floor area is:

TA = 14 (360√3)

= 5040√3

While most spaces of the building are used as rooms, some spaces are wasted because the building is shaped

like a dome. The changes in a ratio of the cuboid volume and the wasted volume as the height ranges from

36m to 54m are invested by using integral.

The calculations to find total building volume and wasted volume for each height from 36m to 54m by using

integration are demonstrated below. (The cuboid volumes are taken from the table 1 on page 2)

In order to find the area of the facade, integral can be applied as it can be used to figure out the total area

under a curve. The equation for the roof structure is y=−h x2

1296+h, the area under this curve will be

A=∫−36

36

(−h x2

1296¿+h)dx .¿

Wasted volume = Total building volume – Cuboid volume

= 150×∫−36

36

(−h x2

1296¿+h)dx−(−25h x3

108+300 x h)¿

= 150h×∫−36

36

( −x2

1296¿+1)dx−(

−25h(12√33)

108+300xh)¿

= 150h×[ −x3

3888+x ] 36

−36−(−1200√3 h+3600√3h)= 150h× [ (−12+36 )−(12−36)]−(2400√3h)= 7200h −¿2400√ 3h

∴ When the height is 36m, Floor Area = 87295.36

= 87295.36

Furthermore, the ratio of wasted volume and the total building volume can be found by using the equation

for total building volume and the equation for wasted volume that I just found. This ratio can be used to

estimate what percentage of space in the building is wasted when the office blocks are fit in a cuboid.

Ratio = Wasted volume ÷Cuboid volume

= (7200h−2400√ 3h¿÷ ¿)

= (7200h−2400√ 3h¿÷ (−25h (12√3

3)108

+300 xh)

= (7200h−2400√ 3h¿÷ (−1200√3h+3600 √3h)

= h(7200−2400√ 3¿÷h (2400√3)

= (7200−2400√ 3¿÷ (2400√3)

= 0.73205081

The ratio of wasted volume and cuboid volume is always 0.73205081 at any height. This shows that about

73 percent of the space inside the building is wasted because the office blocks are only fit into a single

cuboid. The examples shown below are sample calculations for wasted area when the height of the structure

is 36m and 54m.


Vw = 7200h −¿2400√ 3h

= 7200(36) −¿2400(36)√ 3

= 109550.8102

Ratio = 109550.8102 ÷ 259200

= 0.732051


Vw = 7200h −¿2400√ 3h

= 7200(36) −¿2400(36)√ 3

= 164326.2153

Ratio = 164326.2153 ÷ 388800

= 0.732051

Wasted volume = total building volume – cuboid volume

Ratio = wasted area ÷ total building area

Ratio = wasted area ÷ total building area


Table 2: Cuboid volume vs. wasted volume

Height

Integral of

Parabola

Total building

volume Cuboid volume Wasted volume Ratio

36 1728 259200 149649.1898 109550.8102 0.732051

37 1776 266400 153806.1117 112593.8883 0.732051

38 1824 273600 157963.0337 115636.9663 0.732051

39 1872 280800 162119.9556 118680.0444 0.732051

40 1920 288000 166276.8775 121723.1225 0.732051

41 1968 295200 170433.7995 124766.2005 0.732051

42 2016 302400 174590.7214 127809.2786 0.732051

43 2064 309600 178747.6433 130852.3567 0.732051

44 2112 316800 182904.5653 133895.4347 0.732051

45 2160 324000 187061.4872 136938.5128 0.732051

46 2208 331200 191218.4092 139981.5908 0.732051

47 2256 338400 195375.3311 143024.6689 0.732051

48 2304 345600 199532.253 146067.747 0.732051

49 2352 352800 203689.175 149110.825 0.732051

50 2400 360000 207846.0969 152153.9031 0.732051

51 2448 367200 212003.0188 155196.9812 0.732051

52 2496 374400 216159.9408 158240.0592 0.732051

53 2544 381600 220316.8627 161283.1373 0.732051

54 2592 388800 224473.7847 164326.2153 0.732051

As the table shows, although the cuboid volume changes as the height of the structure is altered, the ratio of

cuboid volume and wasted volume is constant for all heights. When the cuboid volume which can be used

for rooms increases, the wasted spaces also increase as the rooms must be cuboid and cannot be flexibly

changed in shapes. Therefore, the floor area and the volume of the rooms can be increased by raising the

height of the structure but the ratio of wasted volume to the cuboid volume cannot be decreased although the

height changes.

(0,0)

(-75,0)

(0,h)(x,y)

(75,0)

Z

(0,h)(x,y)

(0,0) (75,0)(-75,0)

Now, invest changes that occur when the façade is placed on the longer side of the base while the base

remains the same (72m x 150m). As the width of the base changes from 72m to 150m, the equation for this

roof structure is altered.

Diagram 5: The building with a roof structure on the other side

As the height of the structure should be between 50% and 75% of its width, the height can range from 75m

to 112.5m. The diagram below represents a model for the roof structure when the height is 75m.

Diagram 6: The roof structure with the height of 75m

As the roof structure is identical to a parabola being flipped down, the equation of a parabola with a negative

slope is applied for the roof structure.

y = ax2 +bx +c

0 = 752a + 75b + h

0 = (-75)2a – 75b +h

y = h

As the calculation shows, the equation for a roof structure is y = −h

5625x2 + h. [equation 2]

When the height is 75m, the equation for the roof structure is y = −175

x2 + 75.

0 = 752a + 75b + h0 = 752a – 75b + h0 = 150bb = 0

0 = 752a + 75b + h0 = 752a – 75b + h0 = 2(75)2a + 2h5625a = h

a = −h

5625∴y =

−h5625

x2 + h

(0,0)

(-75,0)

(0,h)(x,y)

(75,0)

Z

Since the structure is a building which contains several cubic rooms inside, a cuboid needs to fit inside the

roof structure with maximum volume in order to utilize the space. The diagram 7 demonstrates a cuboid

with maximum volume.

Diagram 7: The cuboid with maximum volume

The dimensions, the volume and the floor area change since its height can vary from 75m to 112.5m.

However, the width represented by z does not change because it is kept constant.

Since the height is known, the three dimensions of the cuboid can be calculated using the equation 1 above.

As the volume is at its maximum, the value of x can be calculated by finding when the derivative of the

volume equation is zero. Then the value of y can also be found by plugging x in the equation 1. Also the

volume can be easily calculated by multiplying all the three dimensions of the cuboid. The two sample

calculations using the heights of 36m and 54m are shown below. The rest of the floor areas for different

heights are calculated in the same way and are illustrated in the table 3 on page 15.

A = 2xy

V = 2xy×72

= 144 x(−h

5625x2 + h)

= −144h

5625x3 + 144xh

= −16h

625x3 + 144xh

The volume of the cuboid differs for each height of the structure and it can be found by the equation above.

Now, by using the equation for the volume, the equation which shows the value of x for different heights can

be found.

V = −16h

625x3 + 144xh

dVdx

=−48h625

x2

+144 h

0 = −48h625

x2

+144 h

144h = 48h625

x2

144 = 48

625x2

x2=1875

x=√1875

x=25√ 3

X is a constant value 25√ 3 at all the heights . Thus, the width of the base of the structure does not change

although its height varies.

The equation which shows the value of y for different heights can be also found by using the equation for

the roof structure (equation 2 above) as x stays constant for all heights

y = −h

5625x2 + h

= −1875h

5625+h

= −h3

+h

= 23h

As the equation shows, y is always two-third of the height of the structure.

The examples below are the sample calculations for the dimensions and the volume of the cuboid, when the

height is 75m and 111m.


The base of the shaded area in the diagram 7 is 2x and the height is y. The equations for the shaded area

and the volume are shown below. This can be applied for any height raging from 75m to 11.2m.

h = 75m V = −16h

625x3 + 144xh

= −48h

25x3 + 10800x

dvdx

= −144

25x2 + 10800

0 = −144

25 x2 + 10800

14425

x2 = 10800

x2 = 1875

x = 25√ 3

y = −hx25625

+ h

= −(25√3)2

75 + 75

= 50


h = 111m V = −16h

625x3 + 144xh

= −1776

625x3 + 15984x

dvdx

= −5328

625x2 + 15984

0 = −5328

625 x2 + 15984

5328625

x2 = 15984

x2 = 1875

x = 25√ 3

y = −37(25√3)2

1752 + 111

∴ x = 25√3 y = 50 Volume = 311769.1m3

∴ x = 25√3 y = 74

Volume = 461418.3m3

V = 2xy×72= (50√3)×50×72= 180000√3= 311769.1

V = 2xy×72= (50√3)×74×72= 266400√3= 461418.3

= 74

Table 3: The dimensions, the volumes and the floor areas for different height

Height X Y Z Volume

75 43.30127 50 72 311769.1

77 43.30127 51.33333 72 320083

79 43.30127 52.66667 72 328396.8

81 43.30127 54 72 336710.7

83 43.30127 55.33333 72 345024.5

85 43.30127 56.66667 72 353338.4

87 43.30127 58 72 361652.2

89 43.30127 59.33333 72 369966.1

91 43.30127 60.66667 72 378279.9

93 43.30127 62 72 386593.7

95 43.30127 63.33333 72 394907.6

97 43.30127 64.66667 72 403221.4

99 43.30127 66 72 411535.3

101 43.30127 67.33333 72 419849.1

103 43.30127 68.66667 72 428163

105 43.30127 70 72 436476.8

107 43.30127 71.33333 72 444790.6

109 43.30127 72.66667 72 453104.5

111 43.30127 74 72 461418.3

This table shows the dimensions, volumes and floor areas for different height, which are calculated by the

way explained in the sample calculations. The values of x stay constant for all heights, but the values of y

changes by two-third as the height increase by one. As the value of y changes, the volume and the floor area

of the cuboid also change.

While most spaces of the building are used as rooms, some spaces are wasted because the building is shaped

like a dome. The changes in a ratio of the cuboid volume and the wasted volume as the height ranges from

75m to 112.5m are invested by using integral. The calculations to find total building volume and wasted

volume for each height from 75m to 112.5m by using integration are demonstrated below.

In order to find the area of the facade, integral can be applied as it can be used to figure out the total area

under a curve. The equation for the roof structure is y=−h x2

1296+h, the area under this curve will be

A=∫−75

75

(−h x2

5625¿+h)dx .¿

Wasted volume = Total building volume – Cuboid volume

= 72×∫−75

75

(−h x2

5625¿+h)dx−(−16h

625x3+144 xh)¿

= 72h×∫−75

75

( −x2

5625¿+1)dx−¿¿)

= 72h×[ −x3

16875+x ] 75

−75−(−1200√3h+3600 √3h)= 72h× [ (−25+75 )−(25−75)]−(2400√ 3h)= 7200h −¿2400√ 3h

Furthermore, the ratio of wasted volume and the total building volume can be found by using the equation

for total building volume and the equation for wasted volume that I just found. This ratio can be used to

estimate what percentage of space in the building is wasted when the office blocks are fit in a cuboid.

Ratio = Wasted volume ÷Cuboid volume

= (7200h−2400√ 3h¿÷ ¿)

= (7200h−2400√ 3h¿÷ (−16h (25√3

3)625

+144 xh)

= (7200h−2400√ 3h¿÷ (−1200√3h+3600 √3h)

= h(7200−2400√ 3¿÷h (2400√3)

= (7200−2400√ 3¿÷ (2400√3)

= 0.73205081

The ratio of wasted volume and cuboid volume is always 0.73205081 at any height. This shows that about

73 percent of the space inside the building is wasted because the office blocks are only fit into a single

cuboid. The examples shown below are sample calculations for wasted area when the height of the structure

is 75m and 112.5m.


Vw = 7200h −¿2400√ 3h

= 7200(75) −¿2400(75)√ 3

= 228230.9

Ratio = 228230.9 ÷ 311769.1

= 0.732051

(4) When the height is 112.5m

Vw = 7200h −¿2400√ 3h

= 7200(112.5) −¿2400(112.5)√ 3

= 337781.7

Ratio = 337781.7 ÷ 461418.3

= 0.732051


Ratio = wasted volume ÷ cuboid volume



Table 4: Cuboid volume vs. wasted volume

Height

Integral of

Parabola

Total Building

Volume Cuboid Volume Wasted Volume Ratio

75 7500 540000 311769.1 228230.9 0.732051

77 7700 554400 320083 234317 0.732051

79 7900 568800 328396.8 240403.2 0.732051

81 8100 583200 336710.7 246489.3 0.732051

83 8300 597600 345024.5 252575.5 0.732051

85 8500 612000 353338.4 258661.6 0.732051

87 8700 626400 361652.2 264747.8 0.732051

89 8900 640800 369966.1 270833.9 0.732051

91 9100 655200 378279.9 276920.1 0.732051

93 9300 669600 386593.7 283006.3 0.732051

95 9500 684000 394907.6 289092.4 0.732051

97 9700 698400 403221.4 295178.6 0.732051

99 9900 712800 411535.3 301264.7 0.732051

101 10100 727200 419849.1 307350.9 0.732051

103 10300 741600 428163 313437 0.732051

105 10500 756000 436476.8 319523.2 0.732051

107 10700 770400 444790.6 325609.4 0.732051

109 10900 784800 453104.5 331695.5 0.732051

111 11100 799200 461418.3 337781.7 0.732051

As the table shows, although the cuboid volume changes as the height of the structure is altered, the ratio of

cuboid volume and wasted volume is constant for all heights. When the cuboid volume which can be used

for rooms increases, the wasted spaces also increase as the rooms must be cuboid and cannot be flexibly

changed in shapes. Therefore, the floor area and the volume of the rooms can be increased by raising the

height of the structure but the ratio of wasted volume to the cuboid volume cannot be decreased although the

height changes.

Although the façade is placed on the longer side of the base and the shape of the building changes, the

ratio of wasted volume to cuboid volume is still 0.732051 and there is much wasted space in the

building if the office blocks form a single cuboid. Therefore, the office blocks should not created in the

shape of a single cuboid in order to maximize the office space.

(0,h)(x,y)

(0,0) (36,0)(-36,0)

(0,h)(x,y)

(0,0) (36,0)(-36,0)

2.5

In order to maximize office space further, the blocks are not in the shape of a single cuboid. The diagrams

below represent the change in the blocks’ shapes.

Diagram 8: Maximizing office space

In the previous model with a single cuboid, about 73 percent of spaces in the building is wasted. However,

by separating each office block and allowing its base width to vary, the office space can be maximized and

the wasted space can be reduced.

The table below shows the dimensions, floor area, and volume of office blocks when the height is 36m.

Table 5: Floor area and volume of office blocks

Heigh

t Y X2 X 2X Floor Area Volume

36 2.5 1206

34.7275

1

69.4550

2 10418.25

26045.6

3

36 5 1116

33.4065

9

66.8131

7 10021.98

25054.9

4

36 7.5 1026

32.0312

3

64.0624

7 9609.37

24023.4

3

36 10 936

30.5941

2

61.1882

3 9178.235

22945.5

9

36 12.5 846

29.0860

8

58.1721

6 8725.824

21814.5

6

36 15 756

27.4954

5

54.9909

1 8248.636

20621.5

9

36 17.5 666

25.8069

8

51.6139

5 7742.093

19355.2

3

36 20 576 24 48 7200 18000

36 22.5 486

22.0454

1

44.0908

2 6613.622

16534.0

6

36 25 396

19.8997

5 39.7995 5969.925

14924.8

1

36 27.5 306

17.4928

6

34.9857

1 5247.857

13119.6

4

36 30 216

14.6969

4

29.3938

8 4409.082 11022.7

36 32.5 126

11.2249

7

22.4499

4 3367.492

8418.72

9

36 35 36 6 12 1800 4500

The table represents floor areas and volumes of each office blocks. As the office blocks are no longer a

single cuboid, they have different width and thus their floor areas and volumes differ from each others’. In

this example, 14 floors can be created inside the building with 2.5 m height for all. The bases of the office

blocks are represented by 2x in the table and they get shorter as the blocks are positioned in higher areas.

Since the base gets shorter, the floor area that can be attained decreases and the volume also decreases.

The total floor area inside the building can be calculated by adding all the floor areas shown in the table and

it will be 98552.36m2.

Similarly, the total volume can be calculated by adding all the volumes shown in the table and it will be

259200m3.

Compared with the previous model with a single cuboid, this model creates an increase in floor area.

Increase in floor area = floor area in this model – floor area in a single cuboid

= 98552.36 - 56118.45

= 42433.92m2


= 259200 - 56118.45

= 12819.09


= 12819.09 ÷ 56118.45

= 0.049456

The ratio of wasted volume to cuboid volume in the model with a single cuboid is 0.73 and much space

in the building is wasted. On the other hand, the ratio in this model of office blocks with different

bases is 0.05 and much less space is wasted. Therefore, if an architect wants to created maximum

office space, he should not create office blocks in the shape of a single cuboid.

Business

International assessment 2