. . . . الــعلــبــه الأكــبــر حــجــمــاً . . . .

A sheet of cardboard 3 ft. by 4 ft. will be made into a box by cutting equal-sized squares from each corner and folding up the four edges. What will be the dimensions of the box with largest volume ?

Let variable x be the length of one edge of the square cut from each corner of the sheet of cardboard.



After removing the corners and folding up the flaps, we have an ordinary rectangular box.




We wish to MAXIMIZE the total VOLUME of the box

V = (length) (width) (height) = (4-2x) (3-2x) (x) .

Now differentiate this equation using the triple product rule, getting

V' = (-2) (3-2x) (x) + (4-2x) (-2) (x) + (4-2x) (3-2x) (1)

= -6x + 4x2 - 8x + 4x2 + 4x2 - 14x + 12

= 12x2 - 28x + 12

= 4 ( 3x2 - 7x + 3 )

= 0

for (Use the quadratic formula



i.e., for


or


But since variable x measures a distance. In addition, the short edge of the cardboard is 3 ft., so it follows that . See the adjoining sign chart for the drivative of V



If

ft.

then

ft.3

is largest possible volume of the box