Question #11582

A 12-ounce soft drink can has volume 21.66 cubic inches. If the top and botton are twice as thick as the sides, find the dimensions (radius and height) that minimize the ammount of metal used in the can.

Expert's answer

V = Pi · r^2 · h

h = V / (Pi · r^2)

m = a · (Ss + 2 · St + 2 · Sb) = a ·

(2 · Pi · r · h + 2 · Pi · r^2 +

2 · Pi · r^2) = a · (2 · Pi · r · V / (Pi ·

r^2) + 4 · Pi · r^2) = a ·

(2 · V / r + 4 · Pi · r^2)

Finding the minimum

of 2 · V / r + 4 · Pi · r^2:

2 · V / r + 4 · Pi · r^2 = 0

V + 2 · Pi · r^3

= 0

r = cubrt(V / (2 · Pi)) = cubrt(21.66 / (2 · 3.14)) = cubrt(21.66 /

(2

· 3.14)) = 1.51 inches

h = V / (Pi · r^2) = 21.66 / (3.14 · 1.51 ·

1.51) = 3.03 inches

Answer: 1.5 inches, 3 inches.

h = V / (Pi · r^2)

m = a · (Ss + 2 · St + 2 · Sb) = a ·

(2 · Pi · r · h + 2 · Pi · r^2 +

2 · Pi · r^2) = a · (2 · Pi · r · V / (Pi ·

r^2) + 4 · Pi · r^2) = a ·

(2 · V / r + 4 · Pi · r^2)

Finding the minimum

of 2 · V / r + 4 · Pi · r^2:

2 · V / r + 4 · Pi · r^2 = 0

V + 2 · Pi · r^3

= 0

r = cubrt(V / (2 · Pi)) = cubrt(21.66 / (2 · 3.14)) = cubrt(21.66 /

(2

· 3.14)) = 1.51 inches

h = V / (Pi · r^2) = 21.66 / (3.14 · 1.51 ·

1.51) = 3.03 inches

Answer: 1.5 inches, 3 inches.

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