A closed rectangular box with base twice as long as it is wide is to be built to contain 72 cubic feet. Determine its dimensions if the material used in the box is to be at a minimum and no allowance is made for thickness of the material.
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L = 2w
V = L * w * h
72 = 2w * w * h
72 = 2w²h
h = 36/w²
Now we can express surface area as function of w:
A = 2 * base area + base perimeter * h
A = 2 * L * w + 2 (L + w) * h
A = 2 * 2w * w + 2 (2w + w) * 36/w²
A = 4w² + 216/w
A' = 8w - 216/w² = 0
8w = 216/w²
w³ = 216/8
w³ = 27
w = 3
L = 2w = 6
h = 36/w² = 4
Dimension of box with least surface area: 6ft x 3ft x 4ft
Mαthmφm
V = L * w * h
72 = 2w * w * h
72 = 2w²h
h = 36/w²
Now we can express surface area as function of w:
A = 2 * base area + base perimeter * h
A = 2 * L * w + 2 (L + w) * h
A = 2 * 2w * w + 2 (2w + w) * 36/w²
A = 4w² + 216/w
A' = 8w - 216/w² = 0
8w = 216/w²
w³ = 216/8
w³ = 27
w = 3
L = 2w = 6
h = 36/w² = 4
Dimension of box with least surface area: 6ft x 3ft x 4ft
Mαthmφm