Calculus I optimization word problem

[kermitt.the.frog]

A box is to be made out of a 6 by 18 piece of cardboard. Squares of equal size will be cut out of each corner, and then the ends and sides will be folded up to form a box with an open top. Find the length (L) , width (W), and height (H) of the resulting box that maximizes the volume.
(Assume that W is < or = to L).

[MAS1]

Originally Posted by kermitt.the.frog

A box is to be made out of a 6 by 18 piece of cardboard. Squares of equal size will be cut out of each corner, and then the ends and sides will be folded up to form a box with an open top. Find the length (L) , width (W), and height (H) of the resulting box that maximizes the volume.
(Assume that W is < or = to L).

W = width
L = length
H = height
V = volume = W*L*H

S = length of the square cutout

W = 6 - 2S
L = 18 - 2S
H = S

V = (6 - 2S)(18 - 2S)(S)
V = (6 - 2S)(18S - 2S^2)
V = 108S - 36S^2 - 12S^2 + 4S^3
V = 4S^3 - 48S^2 + 108S

To maximize the volume, take the derivative and set it equal to 0.

der(V) = 12S^2 - 96S + 108
0 = 12S^2 - 96S + 108
0 = 12(S^2 - 8S + 9)
0 = S^2 - 8S + 9

Use the quadratic formula to solve for S.

S = (8 +- sqrt((-8)^2 - 4(1)(9)))/2(1)
S = (8 +- sqrt(28))/2
S = 4 +- sqrt(7)

S = 4 + sqrt(7) gives a value greater than 6 which cannot be used since the width is only 6. Therefore:

S = 4 - sqrt(7) which is about 1.35425

Plug back in to find W, L, H, and V.

W = 6 - 2(1.35425) = 3.2915
L = 18 - 2(1.35425) = 15.2915
H = 1.35425
Max Volume = 68.16

[jmw106462]

Max Volume = 68.16

[jmw106462]

Max Volume = 68.16