INTERNATIONAL ASSESSMENT 2IB Mathematics HL Year 2<br />Rie Yamada<br />- Modeling a functional building -<br />The buildi...
International assessment 2
International assessment 2
International assessment 2
International assessment 2
International assessment 2
International assessment 2
International assessment 2
International assessment 2
International assessment 2
International assessment 2
International assessment 2
International assessment 2
International assessment 2
International assessment 2
International assessment 2
International assessment 2
International assessment 2
International assessment 2
International assessment 2
International assessment 2
International assessment 2
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International assessment 2

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International assessment 2

  1. 1. INTERNATIONAL ASSESSMENT 2IB Mathematics HL Year 2<br />Rie Yamada<br />- Modeling a functional building -<br />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 <br />Diagram 1: The building with a roof structure<br />(0,0)(-36,0)(0,h)(x,y)(36,0)150m<br />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.<br />Diagram 2: The roof structure with the height of 36m(0,h)(x,y)(0,0)(36,0)(-36,0)<br />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. <br />y = ax2 +bx +c<br />0 = 362a + 36b + h<br />0 = (-36)2a – 36b +h<br />y = h<br />0 = 362a + 36b + h0 = 362a – 36b + h0 = 2(36)2a + 2h1296a = ha = -h12960 = 362a + 36b + h0 = 362a – 36b + h0 = 72bb = 0 <br />∴ y = -h1296x2 + h<br />As the calculation shows, the equation for a roof structure is y = -h1296x2 + h.[equation 1]<br />When the height is 36m, the equation for the roof structure is y = -136x2 + 36.<br />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. <br />Diagram 3: The cuboid with maximum volume<br />(0,0)(-36,0)(0,h)(x,y)(36,0)Z<br />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.<br />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.<br />A = 2xy<br />V = 2xy×150<br />= 300 x(-hx21296 + h)<br />= -25h108x3 + 300xh<br />= -25h108x3 + 300xh<br />The volume of the cuboid differs for each height of the structure and it can be found by the equation above. <br />Now, by using the equation for the volume, the equation which shows the value of x for different heights can be found. <br />V = -25h108x3 + 300xh<br />dVdx= -25h36x2+300h<br />0 = -25h36x2+300h<br />300h = 25h36x2<br />300 = 2536x2x2=432<br />x= 432<br />x=12√3<br />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.<br />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<br />y = -h1296x2 + h<br />= -432h1296+h<br />= -h3+h<br />= 23h<br />As the equation shows, y is always two-third of the height of the structure. <br />The examples below are the sample calculations for the dimensions and the volume of the cuboid, when the height is 36m and 54m.<br />When the height is 36m<br />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.<br />V = 300xy= (12√3) × 24×300= 86400√3= 149649.2h = 36m V = -506x3 + 10800x<br />dvdx = - 25x2 + 10800<br />0 = - 25 x2 + 10800<br />25 x2 = 10800<br />x2 = 432<br />x = 12√3<br />∴ x = 12√3 y = 24 Volume = 149649.2m3y = -hx21296 + h<br />= -(123)236 + 36<br />= 24<br />When the height is 54m<br />V = 2xy×150= (12√3) × 36×150= 64800√3= 224473.8h = 54m V = -252x3 + 16200x<br />dvdx = -752x2 + 16200<br />0 = -752 x2 + 16200<br />752 x2 = 16200<br />x2 = 432<br />∴ x = 12√3 y = 36Volume = 224473.8m3x = 12√3<br />= 36y = -(123)224 + 54<br /> <br />Table 1: The dimensions, the volumes and the floor areas for different height<br />HeightXYZVolumeFloor Area3620.7846124150149649.256118.453720.7846124.66667150153806.156118.453820.7846125.3333315015796362353.833920.784612615016212062353.834020.7846126.66667150166276.962353.834120.7846127.33333150170433.862353.834220.7846128150174590.768589.214320.7846128.66667150178747.668589.214420.7846129.33333150182904.668589.214520.7846130150187061.574824.594620.7846130.66667150191218.474824.594720.7846131.33333150195375.374824.594820.7846132150199532.374824.594920.7846132.66667150203689.281059.985020.7846133.33333150207846.181059.985120.784613415021200381059.985220.7846134.66667150216159.981059.985320.7846135.33333150220316.987295.365420.7846136150224473.887295.36<br />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.<br />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.<br />Diagram 4: How changes in height affect width and height of cuboid<br />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. <br />Similarly, the volume of the cuboid changes as the height of the structure is altered. <br />Diagram 5: How changes in height affect volume of cuboid<br />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. Diagram 4: Floor Areas in the cuboid<br />(0,h)(x,y)(0,0)(36,0)(-36,0)2.5<br />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. <br />Room number = height of cuboid ÷ 2.5 (rejecting the decimal places)<br />Total floor area = (2x × 150) × room number <br />When the height is 36m<br />24m÷2.5m =9.6 The cuboid can have 14 rooms at maximum.<br />As the area for each floor is:<br />A = 2x × 150<br />= 300 (12√3)<br />= 360√3<br />The total floor area is:<br />∴ When the height is 36m, Floor Area = 56118.45TA = 9 (360√3)<br />= 3340√3<br />= 56118.45<br />When the height is 54m<br />36m÷2.5m =14.4 The cuboid can have 14 rooms at maximum.<br />As the area for each floor is:<br />A = 2x × 150<br />= 300 (12√3)<br />= 360√3<br />The total floor area is:<br />∴ When the height is 36m, Floor Area = 87295.36TA = 14 (360√3)<br />= 87295.36= 5040√3<br />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. <br />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)<br />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=-hx21296+h , the area under this curve will be<br />A=-3636(-hx21296+h)dx.<br />Wasted volume = Total building volume – Cuboid volume<br />= 150×-3636(-hx21296+h)dx-(-25hx3108+300xh)<br />= 150h×-3636(-x21296+1)dx-(-25h(1233)108+300xh)<br />= 150h×-x33888+x36-36-(-1200√3h+3600√3h)<br />= 150h×-12+36-(12-36)-(2400√3h)<br />= 7200h -2400√3h<br />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.<br />Ratio = Wasted volume ÷ Cuboid volume<br />= (7200h-2400√3h) ÷(-25hx3108+300xh)<br />= (7200h-2400√3h) ÷(-25h(1233)108+300xh)<br />= (7200h-2400√3h) ÷(-1200√3h+3600√3h)<br />= h(7200-2400√3) ÷h(2400√3)<br />= (7200-2400√3) ÷(2400√3)<br />= 0.73205081<br />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. <br />When the height is 36m<br />Wasted volume = total building volume – cuboid volume<br />Vw = 7200h -2400√3h<br />= 7200(36) -2400(36)√3<br />= 109550.8102<br />Ratio = wasted area ÷ total building area<br />Ratio = 109550.8102 ÷ 259200<br />= 0.732051<br />When the height is 54m<br />Wasted volume = total building volume – cuboid volume<br />Vw = 7200h -2400√3h<br />= 7200(36) -2400(36)√3<br />= 164326.2153<br />Ratio = wasted area ÷ total building area<br />Ratio = 164326.2153 ÷ 388800<br />= 0.732051<br />Table 2: Cuboid volume vs. wasted volume<br />HeightIntegral of ParabolaTotal building volumeCuboid volumeWasted volumeRatio361728259200149649.1898109550.81020.732051371776266400153806.1117112593.88830.732051381824273600157963.0337115636.96630.732051391872280800162119.9556118680.04440.732051401920288000166276.8775121723.12250.732051411968295200170433.7995124766.20050.732051422016302400174590.7214127809.27860.732051432064309600178747.6433130852.35670.732051442112316800182904.5653133895.43470.732051452160324000187061.4872136938.51280.732051462208331200191218.4092139981.59080.732051472256338400195375.3311143024.66890.732051482304345600199532.253146067.7470.732051492352352800203689.175149110.8250.732051502400360000207846.0969152153.90310.732051512448367200212003.0188155196.98120.732051522496374400216159.9408158240.05920.732051532544381600220316.8627161283.13730.732051542592388800224473.7847164326.21530.732051<br />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. <br />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. <br />Diagram 5: The building with a roof structure on the other side<br />(0,0)(-75,0)(0,h)(x,y)(75,0)Z<br />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.<br />Diagram 6: The roof structure with the height of 75m(0,h)(x,y)(0,0)(75,0)(-75,0)<br />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. <br />y = ax2 +bx +c<br />0 = 752a + 75b + h<br />0 = (-75)2a – 75b +h<br />y = h<br />0 = 752a + 75b + h0 = 752a – 75b + h0 = 2(75)2a + 2h5625a = ha = -h56250 = 752a + 75b + h0 = 752a – 75b + h0 = 150bb = 0 <br />∴ y = -h5625x2 + h<br />As the calculation shows, the equation for a roof structure is y = -h5625x2 + h.[equation 2]<br />When the height is 75m, the equation for the roof structure is y = -175x2 + 75.<br />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. <br />Diagram 7: The cuboid with maximum volume<br />(0,0)(-75,0)(0,h)(x,y)(75,0)Z<br />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. <br />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.<br />A = 2xy<br />V = 2xy×72<br />= 144 x(-h5625x2 + h)<br />= -144h5625x3 + 144xh<br />= -16h625x3 + 144xh<br />The volume of the cuboid differs for each height of the structure and it can be found by the equation above. <br />Now, by using the equation for the volume, the equation which shows the value of x for different heights can be found. <br />V = -16h625x3 + 144xh<br />dVdx= -48h625x2+144h<br />0 = -48h625x2+144h<br />144h = 48h625x2<br />144 = 48625x2<br />x2=1875<br />x= 1875<br />x=25√3<br />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.<br />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<br />y = -h5625x2 + h<br />= -1875h5625+h<br />= -h3+h<br />= 23h<br />As the equation shows, y is always two-third of the height of the structure. <br />The examples below are the sample calculations for the dimensions and the volume of the cuboid, when the height is 75m and 111m.<br />When the height is 75m<br />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.<br />V = 2xy×72= (50√3) × 50×72= 180000√3= 311769.1h = 75m V = -16h625x3 + 144xh<br /> = -48h25x3 + 10800x<br />dvdx = -14425x2 + 10800<br />0 = -14425 x2 + 10800<br />14425 x2 = 10800<br />x2 = 1875<br />x = 25√3<br />∴ x = 25√3 y = 50 Volume = 311769.1m3y = -hx25625 + h<br />= -(253)275 + 75<br />= 50<br />When the height is 111m<br />V = 2xy×72= (50√3) × 74×72= 266400√3= 461418.3h = 111m V = -16h625x3 + 144xh<br /> = -1776625x3 + 15984x<br />dvdx = -5328625x2 + 15984<br />0 = -5328625 x2 + 15984<br />5328625 x2 = 15984<br />x2 = 1875<br />∴ x = 25√3 y = 74Volume = 461418.3m3x = 25√3<br />y = -37(253)21752 + 111<br />= 74<br />Table 3: The dimensions, the volumes and the floor areas for different height<br />HeightXYZVolume7543.301275072311769.17743.3012751.33333723200837943.3012752.6666772328396.88143.301275472336710.78343.3012755.3333372345024.58543.3012756.6666772353338.48743.301275872361652.28943.3012759.3333372369966.19143.3012760.6666772378279.99343.301276272386593.79543.3012763.3333372394907.69743.3012764.6666772403221.49943.301276672411535.310143.3012767.3333372419849.110343.3012768.666677242816310543.301277072436476.810743.3012771.3333372444790.610943.3012772.6666772453104.511143.301277472461418.3<br />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.<br />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. <br />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=-hx21296+h , the area under this curve will be<br />A=-7575(-hx25625+h)dx.<br />Wasted volume = Total building volume – Cuboid volume<br />= 72×-7575(-hx25625+h)dx-(-16h625x3 + 144xh)<br />= 72h×-7575(-x25625+1)dx-(-16h(2533)625+144xh)<br />= 72h×-x316875+x75-75-(-1200√3h+3600√3h)<br />= 72h×-25+75-(25-75)-(2400√3h)<br />= 7200h -2400√3h<br />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.<br />Ratio = Wasted volume ÷ Cuboid volume<br />= (7200h-2400√3h) ÷(-16h625x3 + 144xh)<br />= (7200h-2400√3h) ÷(-16h(2533)625+144xh)<br />= (7200h-2400√3h) ÷(-1200√3h+3600√3h)<br />= h(7200-2400√3) ÷h(2400√3)<br />= (7200-2400√3) ÷(2400√3)<br />= 0.73205081<br />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. <br />When the height is 75m<br />Wasted volume = total building volume – cuboid volume<br />Vw = 7200h -2400√3h<br />= 7200(75) -2400(75)√3<br />= 228230.9<br />Ratio = wasted volume ÷ cuboid volume<br />Ratio = 228230.9 ÷ 311769.1<br />= 0.732051<br />When the height is 112.5m<br />Wasted volume = total building volume – cuboid volume<br />Vw = 7200h -2400√3h<br />= 7200(112.5) -2400(112.5)√3<br />= 337781.7<br />Ratio = wasted volume ÷ cuboid volume<br />Ratio = 337781.7 ÷ 461418.3<br />= 0.732051<br />Table 4: Cuboid volume vs. wasted volume<br />HeightIntegral of ParabolaTotal Building VolumeCuboid VolumeWasted VolumeRatio757500540000311769.1228230.90.7320517777005544003200832343170.732051797900568800328396.8240403.20.732051818100583200336710.7246489.30.732051838300597600345024.5252575.50.732051858500612000353338.4258661.60.732051878700626400361652.2264747.80.732051898900640800369966.1270833.90.732051919100655200378279.9276920.10.732051939300669600386593.7283006.30.732051959500684000394907.6289092.40.732051979700698400403221.4295178.60.732051999900712800411535.3301264.70.73205110110100727200419849.1307350.90.732051103103007416004281633134370.73205110510500756000436476.8319523.20.73205110710700770400444790.6325609.40.73205110910900784800453104.5331695.50.73205111111100799200461418.3337781.70.732051<br />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. <br />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. <br />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. <br />(0,h)(x,y)(0,0)(36,0)(-36,0)(0,h)(x,y)(0,0)(36,0)(-36,0)2.5Diagram 8: Maximizing office space<br />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. <br />The table below shows the dimensions, floor area, and volume of office blocks when the height is 36m. <br /> Table 5: Floor area and volume of office blocks<br />HeightYX2X2XFloor AreaVolume362.5120634.7275169.4550210418.2526045.63365111633.4065966.8131710021.9825054.94367.5102632.0312364.062479609.3724023.43361093630.5941261.188239178.23522945.593612.584629.0860858.172168725.82421814.56361575627.4954554.990918248.63620621.593617.566625.8069851.613957742.09319355.23362057624487200180003622.548622.0454144.090826613.62216534.06362539619.8997539.79955969.92514924.813627.530617.4928634.985715247.85713119.64363021614.6969429.393884409.08211022.73632.512611.2249722.449943367.4928418.72936353661218004500<br />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. <br />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.<br />Similarly, the total volume can be calculated by adding all the volumes shown in the table and it will be 259200m3.<br />Compared with the previous model with a single cuboid, this model creates an increase in floor area. <br />Increase in floor area = floor area in this model – floor area in a single cuboid<br />= 98552.36 - 56118.45<br />= 42433.92m2<br />Wasted volume = total building volume – cuboid volume<br />= 259200 - 56118.45<br />= 12819.09<br />Ratio = wasted volume ÷ cuboid volume<br />= 12819.09 ÷ 56118.45<br />= 0.049456<br />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. <br />

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