An open box is to be made from a sheet of card with identical squares cut of the corners to make a box:yx = corner lengthy = box lengthThe card is then folded along the dashed lines to make the box. The aim of this activity is to determine the size of the cut out square which makes the largest volume for a square.Firstly I will find this in a square of y = 10cmTo find the volume I will use this equation-Volume= Width x length x heightCut off (cm)width (cm)length (cm)height (cm)volume (cmï¿½)188164266272344348422416The maximum box volume is made from the cut out square of 2cms. I will now try between 1-2 cm because the highest volume is somewhere between them.
Cut off (cm)width (cm)length (cm)height (cm)volume (cmï¿½)220.127.116.11.166.9241.27.67.
973.036266272The highest value of x is 1.7, but once again I will find a volume even higher in the 10 by 10 box by going into two decimal places.Cut off (cm)width (cm)length (cm)height (cm)volume (cmï¿½)1.616.786.781.
6974.0632The highest value of x is therefore 1.67. I will now go into even smaller numbers between 1.66 and 1.67.Cut off (cm)width (cm)length (cm)height (cm)volume (cmï¿½)1.
0741.6696.6626.6621.66974.074The highest is 1.667.
I will show the graph to show the change in volume for a 10 by 10 box.The maximum value of x to make the highest volume is 1/6 of the length. I can tell this because 10 divided by 6 is 1.
666666667. I will now try a 15 by 15 square. I shall predict the maximum volume of the square by dividing 15 by 6 and this makes 2.5.Cut off (cm)width (cm)length (cm)height (cm)volume (cmï¿½)113131169211112242399324347741965555125633654I will find the maximum value of x between 2 and 3 as it is between them.
Cut off (cm)width (cm)length (cm)height (cm)volume (cmï¿½)2.110.810.82.1244.
4563993243I shall now try between 2.4 and 2.6.Cut off- x (cm)width (cm)length (cm)height (cm)volume (cmï¿½)2.4110.1810.182.
59249.718.104.22.168.6249.704The maximum value of x is 2.5cm.
This shows my prediction was correct. This is that a square box has to be divided by six to find the cut out size in which would give the maximum volume.To help find the highest value of x to make the largest volume, I will use algebra, with aid from textbook sources. I will need to divide the cut out square by the original length of the box.
L= Length of boxY= Length of box- 2×10= 2x+yx = 10/6 or L/6I will substitute these into simple equations:V = (L – 2L/6)(L – 2L/6) L/6I multiplied the 2 in each bracket by L/6 which gives me:V = (L – 2L/6)(L- 2L/6)L/6This makesV = L (L- 2L/6) – “L/6 (L – 2L/6) L/6This is:V = (Lï¿½ – 2Lï¿½/6 – 2Lï¿½/6 – 4Lï¿½)L/6I will multiply the lot by L/6V = Lï¿½/6 – 2Lï¿½/36 – 2Lï¿½/36 + 4Lï¿½/216Now I have put the denominator as 216V = (4Lï¿½ + 36Lï¿½ – 12Lï¿½ – 12Lï¿½)/216This makes40Lï¿½ – 24Lï¿½ / 216This is simplified asV = 2Lï¿½/ 27I will now test this using different size squares.* 10cm by 10cm = 2×10ï¿½/27 = 74.074074* 15cm by 15cm = 2×15ï¿½/27 = 250This proves that the equation is correct.Also there is another way to find out the highest volume shown below. I will show this formula using a 10 by 10 square.
V = x(10-2x)(10-2x) This is minus 2x because of the lengths of the square cut-outs.= x(100-20x-20x+4xï¿½)= 4xï¿½-40xï¿½+100xUsing the gradient function I can work this out. The gradient function is this g = anx n-1.This makes12xï¿½-80x+100 = 0Using axï¿½+bx+c = 0x = – b +/- Vbï¿½-4ac2ax = 80 +/- V-80ï¿½ – (4x12x100)2x12x = 80 +/- V6400 – 480024x = 80 +/- 4024If it were minus then this is the result- 1.
667If it were plus then this is the result- 5 This would be impossible as this would make up the length of the square so minus is correct.I am now going to investigate the open box with a rectangle. I will begin this part of my investigation with a general formula for the volume of the rectangle.The volume of an open box can be expressed like this:V = x (L-2x)(W-2x)This is simplified as:V = x (4xï¿½ + LW – 2Wx – 2Lx)V = 4xï¿½ + 2Lxï¿½ – 2Wxï¿½ + LWxV = x(LW) + 4xï¿½ – xï¿½(2L + 2W)I can now differentiate When dv/dx = 0dv/dx = LW + 12xï¿½ – 2(2L + 2W)0 = 12xï¿½ – (4L + 4W) + LWThis is now a quadratic equation, we can use this formula:x = -b ï¿½ V-(4L + 4W)ï¿½ – 4 x 12 x LW2 x 12This is simplified as:x = 4(L + W) ï¿½ V-(4L + 4W) – 48LW24Multiplying the two brackets together:x = 4(L + W) ï¿½ V16Lï¿½ + 16W2 – 16LW24Factorising:x = 4(L + W) ï¿½ V16 (Wï¿½ + Lï¿½ – LW)24x = 4(L + W) ï¿½ 4V(Wï¿½ + Lï¿½ – LW)24I can simplify by dividing by the common factor of 4:x = L + W ï¿½ VWï¿½ + Lï¿½ – LW6I cannot simplify anymore, but using this I am able to find the value of x which would give me the largest volume of the rectangular box. I shall make predictions to some rectangles and prove them. Firstly I will find the largest volume of the box if the rectangle was 10cm by 15cm.Firstly my prediction:x = 10 + 15 ï¿½ V15ï¿½ + 10ï¿½ – 10x156x = 25 ï¿½ V225 + 100 -1506x = 25 ï¿½ V1756x = 25 + 13.
22875656 = 6.3714594276x = 25 – 13.22875656 = 1.961873907 or 1.962 (3.d.p)6x must be 1.
962 because otherwise 6.371 (3.d.p) is over twice the length. I shall prove that this formula is correct.Cut off (cm)Width (cm)Length (cm)Height (cm)Volume (cmï¿½)11381104211621323943108472456Using the graph I can see that the highest is between 1 and 3.Into decimal points as it is more accurate between 1and 2.
Cut off (cm)Width (cm)Length (cm)Height (cm)Volume (cmï¿½)22.214.171.124.1109.8241.212.
85.82.1131.544The highest is between 1.9 and 2 so I will go into 2 decimal places.Cut off (cm)Width (cm)Length (cm)Height (cm)Volume (cmï¿½)1.
141.93132.0112281.94126.96.36.199132.0255361.95188.8.131.52132.03451.9611.086.081.96132.0381441.9711.066.061.97132.0364921.9811.046.041.98132.0295681.9911.026.021.99132.017396From this the highest is 1.96. I shall go into further decimal places to be certain between 1.96 and 1.97.Cut off (cm)Width (cm)Length (cm)Height (cm)Volume (cmï¿½)1.96111.0786.0781.961132.03821671.96211.0766.0761.962132.03823651.96311.0746.0741.963132.03820341.96411.0726.0721.964132.03811741.96511.076.071.965132.03797851.96611.0686.0681.966132.03778681.96711.0666.0661.967132.03754231.96811.0646.0641.968132.03724491.96911.0626.0621.969132.0368948From this I can tell that 1.962 is the highest value and so my prediction was correct. I shall now try with a rectangle of 20 by 40.First my prediction:x = 20 + 40 ï¿½ V40ï¿½ + 20ï¿½ – 20x406x = 60 ï¿½ V1600 + 400 – 8006x = 60 ï¿½ V12006x = 60 + 34.64101615 = 15.773502696x = 60 – 34.64101615 = 4.226497308 = 4.226 (3.d.p)6x must be 4.226 because 15.774 (3.d.p) would be more than the length.I shall prove this:Cut off (cm)Width (cm)Length (cm)Height (cm)Volume (cmï¿½)13818168423616211523341431428432124153653010515006288613447266710928244876892229396This graph and the table shows that the highest volume achieved would be between 4 and 5:Cut off (cm)Width (cm)Length (cm)Height (cm)Volume (cmï¿½)4.131.811.84.11538.4844.231.611.64.21539.5524.331.411.44.31539.2284.4184.108.40.206537.5364.531114.51534.54.630.810.84.61530.1444.730.610.64.71524.4924.830.410.44.81517.5684.9220.127.116.11509.396From this table, I can see that the highest volume that will be achieved is between 4.2 and 4.3:Cut off (cm)Width (cm)Length (cm)Height (cm)Volume (cmï¿½)4.2131.5811.584.211539.5818444.2231.5611.564.221539.5977924.2331.5411.544.231539.5998684.2431.5211.524.241539.5880964.2531.511.54.251539.56254.2631.4811.484.261539.52310442.731.4611.464.271539.46993242.831.4411.444.281539.4030084.2931.4211.424.291539.322356This shows that the answer is between 4.22 and 4.23:Cut off (cm)Width (cm)Length (cm)Height (cm)Volume (cmï¿½)4.22131.55811.5584.2211539.5986234.22231.55611.5564.2221539.5993164.22331.55411.5544.2231539.599874.22431.55211.5524.2241539.6002864.22531.5511.554.2251539.6005634.22631.54811.5484.2261539.6007014.22731.54611.5464.2271539.60074.22831.54411.5444.2281539.6005614.22931.54211.5424.2291539.600284This shows that 4.226cm needs to be cut off to give the highest volume for a box of a 20 by 40cm rectangle. This proves that my formula is correct.To show that the formula; x = L + W ï¿½ VWï¿½ + Lï¿½ – LW6Is correct by the formula obtained earlier:x = -b ï¿½ V bï¿½ – 4ac2aAlso the formula v = x (L-2x)(W-2x)Using the rectangle 10 by 15:v = x (10-2x)(15-2x)v = x (4xï¿½ – 50x + 150)v = 4xï¿½ – 50xï¿½ + 150xNow differentiate:dv/dx = 4xï¿½ – 50xï¿½ + 150xWhen dv/dx = 00 = 12xï¿½ – 100x + 150Simplified by the common factor of four:0 = 3xï¿½ – 25x + 37.5This is now ready for the formula:x = 25 ï¿½V25ï¿½-4x3x37.52x3x = 25 ï¿½V1756x = 25 + 13.22875656= 6.3714594276x = 25 – 13.22875656 = 1.96187390761.962 (3.d.p) is correct as 6.371 (3.d.p) would be too large. This shows that my formula is correct. I shall show this with the 20 by 40 rectangle:v = (20-2x)(40-2x)v = (4xï¿½-120x+800)v = 4xï¿½ – 120xï¿½ + 800xNow differentiate:dv/dx = 4xï¿½ – 120xï¿½ + 800xWhen dv/dx = 00 = 12xï¿½ – 240x + 800Divide by the common factor of four:0 = 3xï¿½ – 60 + 200Now I can use the formula:x = 60 ï¿½V60ï¿½ – 4x3x2002x3x = 60ï¿½V3600 – 4x3x2006x = 60 ï¿½/12006x = 60 + 34.64101615 = 15.773502696x = 60 – 34.64101615 = 4.22649730864.226 (3.d.p) is correct because 15.774 (3.d.p) is too large to make the box.These results confirm that the formula is correct:x = L + W ï¿½ VWï¿½ + Lï¿½ – LW6This formula shows the size of the cut off which gives the highest possible volume for a rectangular box.