precalculus
posted by kim .
a rectangular piece of cardboard measuring 14 inches by 27 inches is to be made into a box with an open top by cutting squares of equal size from each corner and golding up the sides . let x represent the length of a side of each square. for what value of x will the volume be a maximum? round to two decimal places.

v = x(142x)(272x)
dv/dx = 2(6x^2  82x + 189)
dv/dx=0 when x = (41 ± √547)/6 = 2.93, 10.73
knowing the shape of cubics, you should have no trouble "differentiating" the max from the min. 
length = z = 27 2x
width = y = 14  2x
v = (272x)(142x)x
v = ( 378  82 x + 4 x^2 ) x
v = 378 x 82 x^2 + 4 x^3
max when dv/dx = 0
dv/dx = 378  164 x + 12 x^2 = 0
6 x^2  82 x + 189 = 0
x = [ 82 +/ sqrt (67244536) ]/12
x = [82 +/ 46.8 ]/12
x = 10.7 too big, negative width
x = 2.93 inches