an open box is to be made from a square piece of cardboard whose sides are 19 inches long,by cutting squares of equal size from the corners and bending up the sides. Determine the size of the aquare that is to be cut out so that the volume may be a maximum.

Answer 1

Let the size of the cut-out square be x inches
so the base of the box is (19-2x) by (19-2x) and the height is x
V = x(19-2x)^2
= x(361 - 76x + 4x^2)
= 4x^3 - 76x^2 + 361x

dV/dx = 12x^2 - 152x + 361
= 0 for a max of V
x = (152 ± √5776)/24
= (152 ± 76)/24 = 19/2 or 19/6
but 0 < x < 19/2
so x = 19/6
the square to be cut out should be appr 3.17 inches by 3.17 inches.

(notice our quadratic 12x^2 - 152x + 361 = 0 factors to
(2x - 19)(6x - 19) = 0
x = 19/2 or x = 19/6