1determine the value of h,r and hence calculate the ratio of h/r when the total surface area of each tin is minimum.here,h cm denotes the height and r cm the radius of the tin.2the t0p and bottom pieces of height h cm are cut from square-shape aluminium sheets.determine the value for r,h and hence calculate the ratio h/r so that the total area of the alumium sheets used for making the tin is minimum.(refer to the diagram the aluminium sheets used for making the tin is minimum.3investigate cases where the top and bottom surfaces are cut from;i)equilateral triangle.ii)regular hexagon.find the ratio of hj/r for each cases.further investigationinvestigate cases where the top and bottom faces of the tin are being cut from aluminium sheets consisting shapes of regular polygons.from the results of your investigation,what conclusion can you derive frrom the relationship of the ratio h/r and the number of sides of a regular polygon?
1Here is the first partWhat is the ratio of the height to the radius of a can that has minimum surface area for a given volume? r radius can h height can V volume S surface area V πr?h S 2πr? + 2πrh Solve for hV πr?h h V/(πr?) Substitute into SS 2πr? + 2πrh 2πr? + 2πr{V/(πr?)} 2πr? + 2V/r Take the derivative of S to find critical valuesS 2πr? + 2V/r dS/dr 4πr - 2V/r? 0 4πr 2V/r? 2πr V/r? r? V/(2π) r {V/(2π)}^? Take the second derivative of S to find the nature of the critical valuesdS/dr 4πr - 2V/r? d?S/dr? 4π + 4V/r? 0 since both terms are positive This implies a relative minimum, which is what we wantSolve for hh V/(πr?) 2V/(2πr?) (2/r?)(V/2π) (2/r?)(r?) 2r h 2r Ratio of h/r h/r 2r/r 2
