An open box is to be made out of a 6-inch by 20-inch piece of cardboard by cutting out squares of equal size f

An open box is to be made out of a 6-inch by 20-inch piece of cardboard by cutting out squares of equal size from the four corners and bending up the sides. Find the dimensions of the resulting box that has the largest volume.
Dimensions of the bottom of the box: x
Height of the box:

Base of box will be (20-2f) by (6-2f). Height will be equal to f.

Volume (V) = area of base * height
V = ( 20 - 2f)*(6-2f)*f
Opening the brackets by multiplying:
V = (120 -12f -40f + 4f^2)*f
V = 120f -52f^2 + 4f^3
V = 4f^3 - 52f^2 +120f
Differentiating Volume with repsect to height f:
dV/df = 12f^2 -104f +120
For a maximum value:
dV/df = =0
Divinding by 4:
3f^2 - 26f + 30 =0
Using f = [-b+/-(b^2 -4ac)^0.5] /2a
f = [26+/- (26^2 - 4*3*30)^0.5]/6
f = 26 +/- 17.78 / 6
f = 7.296 or 1.371


For maximum volume: dV/df = 0 (differentiation)
So dV/df = 0 = 12f^2 - 104f + 120

Side (6-2f) shows that f cannot be 7.296
So f = 1.371
Bottom sides : (20 - 2f) = 17.258 inches by 3.258 inches (6 - 2f).
Height of box will be 1.371 inches.
Volume will be: 17.258 * 3.258 * 1.371= 77.1 inch cubed