Cardboard open top box

Question

A rectangular piece of cardboard measuring 120 cm by 80 cm is used to create an open-topped box by cutting out congruent squares from all four corners. Determine the dimensions of the square that must be cut from each corner in order for the volume to equal 38 500 cm?

State the dimensions of all possible boxes that have a volume of 38 500 cm.

If you want the box to have a volume greater than 38 500 cm, what side lengths of the cut-out square are permissible?

Answer 1

x(120-2x)(80-2x)=3850= dLw

4x^3-400x^2+9600x-3850=0

2x^3-200x^2+4800x-1925=0

1 or 3 positive real answers by descartes rule of signs

x= 5 cm so (x-5) is a factor

use long or synthetic division to get the quadratic factor

(x-5)(2x^2-190x+3850)= 0

2(x-5)(x^2 -95x+1925) = 0

x = 95/2 +/-(1/2)sqr(95^2 -4(1925))

= 47.5 +/-.5sqr(9025-7700)

=47.5+/-.5sqr1275

=47.5+/-5sqr13

= about 29.5, 65.5,

= about 29.5. reject 65.5 as 80-2x then is negative for width of the box

5 cm < x < about 30 cm for volume > 38,500 cm^3

Answer 2

Let x = length of a side of each square

V = x(120 - 2x)(80 - 2x), 0 < x < 40

V = x(4x² - 400x + 9600)

V = 4x³ - 400x² + 9600x

4x³ - 400x² + 9600x - 38500 = 0

By trial and error, 5 is a root. Dividing by x - 5 gives (x - 5)(4x² - 380x + 7700) = 0

4(x - 5)(x² - 95x + 1925) = 0

Use the quadratic formula to find the remaining roots.