Folded boxes (a). Squares with sides of ?length x ? are cut out of each corner of a rectangular piece of cardboard measuring 3 ft by 4 ft. The resulting piece of cardboard is then folded into a box without a lid. Find the volume of the largest box that can be formed in this way. (b). Suppose that in part (a) the original piece of cardboard is a square with sides of ? length ??. Find the volume of the largest box that can be formed in this way. (c). Suppose that in part (a) the original piece of cardboard is a rectangle with sides of ? ?length ?? an? d? . Holding ?€ fixed, find the size of the c?orner squares x ? that maximizes the volume ? of the box as L ? ? ?. (?Source: Mathe? matics Teacher, November 2002)
Solution 26E Step 1: (a) onsider abcd is the rectangular sheet of paper such thatab andcd are the lengths and ac andbd are the widths. If a square with sides of length x is cut from each corner, then the new dimensions of the sheet measuring 3ft by 4ft will be in length and in width. Consider the following figure Step 2: Consider V is the volume of the resulting piece of cardboard that is rectangular solid without lid, which is obtained by folding the rectangular sheet after cutting out the squares. The dimensions for computing the volume will be x,32x,42x Therefore the volume will be, V = x(32x)(42x) =(3x2x )(42x) 2 3 V =12x14x +4x …(1) dV Find the critical value of x by equating the derivative dx to zero. dV 2 dx =1228x+12x ….(2) After solving this quadratic equation we get x=1.76,x=0.56 Substitute both the values in the second derivative to find which maximizes the volume.. If the second derivative is positive at a point then it will have minima at that point otherwise maxima. If x = 1.76, then second derivative will be, d V =28+24 (since d x = nx n1) dx2 dx d V dx2 (x=1.76)28+24(1.76) =14.24 Since forx = 1.76, the second derivative is a positive value, the volume is not maximized at this point. Now we check for x=0.56 , then second derivative will be, d V d n n1 dx2 =28+24 (since dxx = nx ) d V dx2 (x=0.56)28+24(0.56) =-14.56 Since for x = 0.56, the second derivative is a negative value, the volume is maximized at this point. Therefore substitute x = 0.56 into V , V =12x14x +4x 2 3 V = 12(0.56)14(0.56) +4(0.56) 3 =3.03 Thus the volume of the largest box that can be formed is 3.03 cubic feet.