What is the volume of the box in the following case:
A piece of cardboard is rectangular in shape. Squares of side 2 cm are cut from each of the corners of the cardboard and the remaining part is folded to form an open rectangular box. Given that the perimeter of the cardboard (before the corners are cut) is 30cm and the surface area of the box that is formed is 40cm^2, find the volume of the box.
Let the dimensions of the original piece of cardboard be x and y. The perimeter of this piece is 30 cm.
=> x + y = 15
Squares of side 2 cm are cut from each corner of the piece and the rest is folded to form a rectangular box with surface area 40 cm^2.
The surface area of the rectangular box is (x - 4)(y - 4) + 2*(x-4)*2 + 2*(y-4)*2 = 40
=> xy - 4y - 4x + 16 + 4x - 16 + 4y - 16 = 40
=> xy = 56
To determine x and y the equations xy = 56 and x + y = 15 have to be solved.
=> x(15 - x) = 56
=> 15x - x^2 = 56
=> x^2 - 8x - 7x + 56 = 0
=> x(x - 8) - 7(x - 8) = 0
=> (x - 7)(x - 8) = 0
=> x = 7 and x = 8
y = 8 and y = 7
The volume of the box is 3*4*2 = 24 cm^3
An alternative, albeit more complex, solution to the previous answer using the quadratic equation as follows:
Once you have the two equations xy = 56 and x + y = 15 and reach this point 15x - x^2 = 56 in the solution, you may express this in quadratic form (ax^2+bx+c=0) as
x^2 - 15x + 56 = 0
and solve using the quadratic equation
x = [-b +- sqrt( b^2 - 4ac)] / 2a
x = [15 +- sqrt(225 - 224)] / 2
x = [15 +- 1] / 2
x = 16/2 and 14/2 or 8 and 7
volume is calculated after taking off the 2cm squares, in other words, subtracting 4cms from x and y.
8-4 = 4 and 7-4 = 3 so Volume = 4 x 3 x 2(height) = 24cm^3