Abox is to be constructed from a sheet of cardboard that is 10 cm by 60 cm by cutting out squares of length x by x from each corner and bending up the sides. what is the maximum volume this box could have?

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

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.

step-by-step explanation:

the answer is 128