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calc help!!!!!!!!!!!?
A box has a square base of side x and height y.
(a) Find the dimensions x,y for which the volume is 6 and the surface area is as small as possible.
(b)Find the dimensions for which the surface area is 8 and the volume is as large as possible.
without cover i think
1 Answer
- Ossi GLv 79 years agoFavorite Answer
with or without cover?
a) with cover:
V = x^2*y = 6 --->y = 6/x^2
A = 2x^2+ 4xy
A = 2x^2 + 4x(6/x^2)
A = 2x^2 + 24/x
set A' = 0 to find the minimum:
A' = 4x - 24/x^2 = 0
4x = 24/x^2
x = 6/x^2
x^3 = 6
x = 6^(1/3) = 1.817
y = 6/x^2 = 6/(6^(2/3) = 1.817
all dimensions are equal--> cube.
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b)
A = 8 = 2x^2 + 4xy ---> y = (8 - 2x^2)/(4x)
V = x^2*y
V = x^2(8-2x^2)/(4x)
V = x(8- 2x^2)/4
V = (8x - 2x^3)/4
set V' = 0 to find the maximum
V' = 1/4(8 - 6x^2) = 0
8 = 6x^2
x^2 = 8/6
x = √(8/6) = 1.154
y = (8-2x^2)/(4x)
y = (8- 16/6)/(4√(8/6))
y = (16/3)/ 4.619
y = 1.1546
---> again a cube with side length 1.1546
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