Answer:
Dimension of bottom =Length x Breadth
= 12.619 inch x 4.619 inch
≈ 13 inch x 5 inch (To nearest whole number)
Height of box = 1.6905 inch
≈ 2 inch
Step-by-step explanation:
Length of the cardboard = 16 inch
Breadth of the cardboard = 8 inch
Since the cardboard has four corners, we would cut out four (4 ) squares.
Assuming the length of the square =y
length of open box = 16-2y----------------------------------- (1)
breadth of open box = 8-2y-------------------------------------(2)
height of open box = y
Volume of box = Length x breadth x height
= (16-2y) x ( 8-2y) x y
= ( 128-48y + 4y²)y
= 128y-48y² + 4y³
To maximize the volume of the box, V(y); dV/dy =0
dV/dy = 128- 96y + 12y²
12y²-96y+128 = 0
a = 12 , b= -96 , c= 128
y= [-b± √(b²-4ac)]/2a
= [96 ±√(96²- (4 x 12 x128)]/2 (12)
y= 6.3094 or y =1.6905
Substituting both values of y into (1) and (2)
When y = 6.3094
Length of box = 16- 2(6.3094)
= 3.3812 inch
Breadth of box =8- 2 (6.3094)
= -4.6188 inch
Since the breadth cannot be negative, this value of y is therefore an extraneous root.
So the correct value of y is the smaller of the two roots of equation,
y = 1.6905 inch
Substituting both values of y into (1) and (2)
When y = 1.6905
Length of box = 16- 2(1.6905)
= 12.619 inch
≈ 13 inch
Breadth of box =8- 2 (1.6905)
= 4.619 inch
≈ 5 inch
So the Dimension of the bottom (to nearest whole number):
Length x breadth = 13 inch x 5 inch
Height of the box = 1.6905 inch
≈ 2 inch
