What are the dimensions of the box of largest volume I can make this way? and what is its volume?

Question

Need to make an open rectangular box from a 12- by 14-cm piece of cardboard by cutting congruent squares from the corners and folding up the sides.

Answer 1

let x = the size of the squares cut from the corners
V = (12 - 2x)(14 - 2x)(x)
V = 4x(6 - x)(7 - x)
V = 4(42x - 13x^2 + x^3)
dV/dx = 4(42 - 26x + 3x^2 = 0 for min/max
x = (26 ± √(676 - 504))/6
x = (26 ± √172)/6
x = 2.1475 cm
L = 9.7050 cm
W = 7.7050 cm
V = 160.58 cm^3

. . x . . . . . V
2.1000 160.524
2.1475 160.5836612
2.2000 160.512

Answer 2

The dimensions of the box are:
14-2y, 12-2y, y
let x = 2y
The volume 2V=x*(12-x)(14-x) = x^3-26x^2+168 x
dV / dx= 3x^2 - 52x+168 = 0
X=(52+ROOT(688))/6 or (52+ROOT(688))/6
x= 13.038292349534667101562739998424
or x= 4.2950409837986662317705933349093
But the first value rejected
So y=2.1475204918993331158852966674545
12-x=7.704959016201333768229406665091
14-x=9.704959016201333768229406665091
are the dimensions of the box
The volume is:
V=321.16732239702875736343509919902/2
=160.5836611985143786817175495995

Answer 3

Let x be the side of cut-out square.
Volume = x (12-2x)(14-2x)

Expand that, get the first derivative of V and set to zero.
That would be the optimal value of x.

Need help or more explanations?
Note: Observe that 0< x < 6.
The acceptable solution for V' = 0 is x = 8/3.
Thus, V = 4160/27 cm³ (154.074)