To solve this problem, let us first state the variables. Let us say that:
l = length of the card board = 60 cm
w = width of the card board = 10 cm
h = the length of the cut being done on both sides = x
We know that the volume for a box is calculated using the formula:
V = l w h
However this is not yet complete, since the h or x side was taken from the length of l and w. Since two sides are being taken up, so we subtract 2 x from l and w. Therefore:
V = (l – 2 x) (w – 2 x) (x)
V = (60 – 2 x) (10 – 2 x) (x)
V = 600 x – 120 x^2 – 20 x^2 + 4 x^3
V = 4 x^3 – 140 x^2 + 600 x
The maxima points are obtained by taking the 1st derivative of the equation and setting up dV / dx = 0, therefore:
dV / dx = 12 x^2 – 280 x + 600
0 = 12 x^2 – 280 x + 600
x^2 – (280/12) x = - 50
By completing the square:
x^2 – (280/12) x + (78,400/576) = - 50 + (78,400/576)
(x – (280/24))^2 = 86.11
x – (280/24) = ± 9.28
x = (280/24) ± 9.28
x = 2.39, 20.95
Since x cannot be bigger than w, therefore the right answer is:
x = 2.39 cm
Therefore the maximum volume is:
Vmax = (60 – 2 * 2.39) (10 – 2 * 2.39) (2.39)
Vmax = 688.91 cm^3
