EXTENSION 1 GENERALLY : If the volume to be used is V then: V = h = V πr2h so πr2 The surface area S = 2πr2 + 2πrh = 2πr2 + 2πr×V πr2 = 2V r 1 2πr2 + – dS = 2Vr 2 = 0 for min S 4πr – – dr 2V = 0 4πr – r2 = 2V 4πr r2 r3 = V 2π So r = V and diameter = 2r = 2× V = = V ⅓ ⅓ ⅓ ×2⅔ (2π)⅓ (2)⅓ (π)⅓ π⅓ and h = V πr2 = V × (2π)⅔ π V⅔ = V ⅓ ×2⅔ π⅓ The ideal can which uses the least amount of metal has the same HEIGHT as its DIAMETER!
EXTENSION 1 GENERALLY : If the volume to be used is V then: V = πr2h so h = V πr2 The surface area S = 2πr2 + 2πrh = 2πr2 + 2πr×V πr2 = 2πr2 + 2V r – 1 dS = 4πr – 2Vr – 2 = 0 for min S dr 4πr – 2V = 0 r2 4πr = 2V r2 r3 = V 2π So r = V ⅓ and diameter = 2r = 2× V ⅓ = = V ⅓ ×2⅔ (2π)⅓ (2)⅓ (π)⅓ π⅓ and h = V πr2 = V × (2π)⅔ π V⅔ = V ⅓ ×2⅔ π⅓ The ideal can which uses the least amount of metal has the same HEIGHT as its DIAMETER!
5c_the_ideal_can__whiskas_.docx