Asked by Jacob
If an open box is made from a tin sheet 7 in. square by cutting out identical squares from each corner and bending up the resulting flaps, determine the dimensions of the largest box that can be made. (Round your answers to two decimal places.)
Height:
Length:
Width:
Height:
Length:
Width:
Answers
Answered by
Reiny
base = 7- 2x
height = x
volume = x(7-2x)^2
= 49x - 28x^2 + 4x^3
d(volume)/dx = 49 - 56x + 12x^2
= 0 for a max volume
12x^2 - 56x + 49 = 0
x = (56 ± √784)/24
= (56 ± 28)/24 = 3.5 or 7/6 or 1.1666...
but clearly x < 3.5 or we have cut the whole base away.
base is 7 - 2(7/6) = 14/3 by 14/3
and the height is 7/6
round to your required decimals
height = x
volume = x(7-2x)^2
= 49x - 28x^2 + 4x^3
d(volume)/dx = 49 - 56x + 12x^2
= 0 for a max volume
12x^2 - 56x + 49 = 0
x = (56 ± √784)/24
= (56 ± 28)/24 = 3.5 or 7/6 or 1.1666...
but clearly x < 3.5 or we have cut the whole base away.
base is 7 - 2(7/6) = 14/3 by 14/3
and the height is 7/6
round to your required decimals
Answered by
Damon
new length = 7 - 2 h
height = h
volume = (7-2h)(7-2h)(h)
v = (49 -28 h + 4 h^2)h
so
v = 4 h^3 -28 h^2 + 49 h
dv/dh = 0 for max or min
dv/dh = 12 h^2 -56 h + 49 = 0
(6h -7)(2 h -7) = 0
h = 7/6 or h = 7/2
if h = 7/2, the box has zero bottom
so the answer is h = 7/6
7 - 2(7/6) = 7-7/3 = 14/3 = length and width
height = h
volume = (7-2h)(7-2h)(h)
v = (49 -28 h + 4 h^2)h
so
v = 4 h^3 -28 h^2 + 49 h
dv/dh = 0 for max or min
dv/dh = 12 h^2 -56 h + 49 = 0
(6h -7)(2 h -7) = 0
h = 7/6 or h = 7/2
if h = 7/2, the box has zero bottom
so the answer is h = 7/6
7 - 2(7/6) = 7-7/3 = 14/3 = length and width
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