A closed aluminum can (right circular cylinder with top and bottom) is required to hold 128π cubic inches of liquid. Find the radius and the height of the can that require the least amount of aluminum.
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r = radius of can
h = height of can
V = π r² h = 128π
r² h = 128
h = 128/r²
Surface area of can:
A = 2πrh + 2πr²
Now if we substitute h with 128/r², then we can express surface area as function of r only
A = 2πr(128/r²) + 2πr²
A = 2π (128/r + r²)
To minimize, we find value of r where A' = 0 and A'' > 0
A' = 2π (−128/r² + 2r) = 0
2r = 128/r²
2r³ = 128
r³ = 64
r = 4
A'' = 2π (256/r³ + 2) > 0 when r = 4
h = 128/r² = 128/16 = 8
Surface area is minimized when
radius = 4 inches
height = 8 inches
h = height of can
V = π r² h = 128π
r² h = 128
h = 128/r²
Surface area of can:
A = 2πrh + 2πr²
Now if we substitute h with 128/r², then we can express surface area as function of r only
A = 2πr(128/r²) + 2πr²
A = 2π (128/r + r²)
To minimize, we find value of r where A' = 0 and A'' > 0
A' = 2π (−128/r² + 2r) = 0
2r = 128/r²
2r³ = 128
r³ = 64
r = 4
A'' = 2π (256/r³ + 2) > 0 when r = 4
h = 128/r² = 128/16 = 8
Surface area is minimized when
radius = 4 inches
height = 8 inches
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Make the height of the can equal to the twice the radius of the can.