Open box, 21 ft by 56 ft rectangular sheet of metal, equal size from four corners. What's maximum volume in ft^3

Question

An open box is to be made from a 21 ft by 56 ft rectangular piece of sheet metal by cutting out squares of equal size from the four corners and bending up the sides. Find the maximum volume that the box can have.

Round your answer to the nearest integer.

Answer 1

21 by 56 feet

cut off corners of equal length leaving a box that's (21-2x) by (56-2x) = volume of (1176 -154x + 4x^2)(x) = 1176x -154x^2 + 4x^3

take the derivative and set equal to zero

V' = 12x^2 -308x + 1176 = 0

3x^2 -77x + 294 = 0

factor, complete the square or use the quadratic formula

x = 77/6 + or - (1/6)sqr(77^2 -12(294))

= (77-49)/6 = 14/3 = 4 2/3

dimensions are 21-2(14/3) by 56-2(4 2/3) by 4 2/3 feet

=11 2/3 by 46 2/3 by 4 2/3 feet

= about 11.7 by 46 2/3 by 4.7 feet

V' = 3x^2 -77x + 294 = 0

3(x^2 -77x/3 + 98) = 0

3(x-14/3)(x-21) = 0

x-14/3 = 0, x-21=0

x= 14/3,

(ignore the other solution: x=21, as it gives a negative dimension, -21 by 14 by 21 feet and the minimum volume: (-21(14)(21) = -6,174)

maximum volume = V = lwh = (11 2/3)(46 2/3)(4 2/3) = (35/3)(140/3(14/3) = 68600/27 = 2,540.74 ft^3

graph V(x) = (21-2x)(56-2x)(x) and it has two turning points or relative extrema, a local max and a local min. The local maximum is the maximum volume, given the 3 dimensions have to be positive numbers.

Answer 2

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