# Applied Calculus

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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:

• Applied Calculus -

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

• Applied Calculus -

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

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