842 words - 3 pages

Mathematical Modelling: Can of Drink1.)Consider a drink can of capacity 330ml. Draw a diagram, or a number of diagrams which show all the features of the can.Height: 11.5 cmRadius: 3.25 cmDiameter (top): 5.42 cmDiameter (middle): 6.5 cmDiameter (bottom): 5 cmCapacity: 330mlVcan: 330 cm"2.)Use your diagrams of part 1 to help you choose a simple three dimensional shape which models this can and define suitable variables for this shape.- I have used a cylinder to model a regular can of drink since it has similar features as a can, one of them being that one can unfold the cylinder into a rectangle (the sides) and two circles (top and bottom) which one can do with a can as well.Solid Cylinder:Area: A = 2πrh + 2πr"Volume: V = (Area of base) x heightV(r) = πr"h1ml = 1cm"Variables are r and hr = radiush = height3.)By considering the volume of this shape, establish a relationship between the variables defined in part 2.h = 11.5 cm ; r = 3.25 cm ; d = 6.5 cmV(r) = πr"h h =- I am using the relationship of the Volume Formula with the variables r and h to eliminate h as an unknown variable. By doing this step I can substitute h with the rearranged Formula, meaning that only r is left as an unknown variable.4.)Find a function, S, in terms of one of the variables defined in part 2, where S is the total Surface Area of the can.330cm" = V = 2πr"hA = 2πrh + 2πr" = 2πr(r + h)- I substituted h with the equation I derived from the Volume formula.S(r) = 2πr" + 2πr ( )S(r) = 2πr" + S(r) = 2π (3.25) " +S(r) = 269.443 cm"- to obtain a suitable equation S, I first split up the cylinder in 3 different pieces. 2 circles for the top and bottom and a rectangle for the sides:- after splitting the cylinder into different parts, I could add the different area equations of the different shapes together and by adding them together I get an equation for the cylinder, the equation S.5.)Use differentiation to find the values of your variables which minimize the surface area of the can.To find the minimum value, one should use differentiation (taking the derivative), since stationary points (min. and max.) of the function S occur when S = 0S (r) =0 =- since it is a cubic function one will get 3 different results of which two will be...

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