Area & Volume Exploration &#150; Component proportional changes
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Introduction
Area & Volume Exploration – Component proportional changes
Question 2: Suppose that you are required to design an open steel tray from a sheet of steel that measures 120cm by 80cm by cutting squares from each corner, and folding to form a tray. What size should the squares be cut from each corner of the sheet so that the maximum volume is obtained for the tray.
Height = x cm
Width = 80cm  2x cm
Length = 120cm – 2 x cm
The task is to find the optimal size of the squares that need to be cut out of the corners in order to find the maximum obtainable Volume inside the tray once it has been folded. To work this out there are a number of different methods for doing this. One method is Trial and Improvement.
Trial and Improvement
Height (Value of xcm)  2 xcm  Length (cm) (120cm – 2 x cm)  Width (cm) (80cm  2x cm)  Volume (cm3) (120x  2 x 2)×(80  2x)  +/ 
5  10  110  70  38’500  
7.5  15  105  65  51’187.5  + 
10  20  100  60  60’000  + 
12.5  25  95  55  65’312.5  + = 
15  30  90  50  67’500  + 
17.5  35  85  45  66’937.5   
Middle
+
15
30
90
50
67’500
+
16
32
88
48
67’584
+
=
17
34
86
46
67’252

The Volume starts to decrease after the value of 16 for x. So we know that the Maximum volume lies between 15 and 17. We can now continue the trial and improvement table working only between 15 and 17.
Height (Value of xcm)  2 xcm  Length (cm) (120cm – 2 x cm)  Width (cm) (80cm  2x cm)  Volume (cm3) (120x  2 x 2)×(80  2x)  +/ 
15  30  90  50  67’500  
15.25  30.5  89.5  49.5  67’561.3125  + 
15.5  31  89  49  67’595  + 
15.65  31.3  88.7  48.7  67’603.1485  + = 
15.75  31.5  88.5  48.5  67’602.9375   
The table shows that the volume is at its greatest at 15.65 and decreases after this value. For increased accuracy we can continue the trial and improvement table working between the x values of 15.50 and 15.8.
Height (Value of xcm)  2 xcm  Length (cm) (120cm – 2 x cm)  Width (cm) (80cm  2x cm)  Volume (cm3) (120x  2 x 2)×(80  2x)  +/ 
15.500  31  89  49  67’595  
15.525  31.05  88.95  48.95  67’597.44131  + 
15.550  31.1  88.9  48.9  67’599.1155  + 
15.575  31.15  88.85  48.85  67’600.52294  + 
15.600  31.2  88.8  48.8  67’601.664  + 
15.625  31.25  88.75  48.75  67’02.53.906  + 
15.650  31.3  88.7  48.7  67’603.1485  + 
15.675  31.35  88.65  48.65  67’603.49269  + 
15.700  31.4  88.6  48.6  67’603.572  + = 
15.725  31.45  88.55  48.55  67’603.38681   
The optimal value of xto create a maximum volumehas now been narrowed down to between 15.675 and 15.725.
Conclusion
There are other methods that could be used to determine the maximum volume and the optimal value of X.One method could be differential calculus. This is applying the rules of differentiation to the object in question. This would be a quicker and more accurate method to determine the maximum volume and vale of X.
This investigation could be useful in engineering projects where a product must be produced with the maximum efficiency. This could mean being produced so that the volume inside is at its maximum possible and the wastage of material is at its least. Differentiation can also be used to determine dimensions of building sites or fields etc.
This exercises a very useful task that is widely used throughout many types of engineering and construction.
This student written piece of work is one of many that can be found in our GCSE Open Box Problem section.
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