Volume of box?

Question

Heloise designed a cubic box for shipping paper. The height of the box was acceptable, but the paper would not fit into the box. After increasing the length by 0.5 in. and the width by 6 in. the area of the bottom was 119 in.^2 and the paper would fit. What was the original volume of the cubic box?

Answer 1

(x + 0.5)(x + 6) = 119
x^2 + 6.5x + 3 = 119
x^2 + 6.5x - 116 = 0
x = (- 6.5 ± √(42.25 + 464))/2
x = (- 6.5 ± √506.25)/2
x = (- 6.5 ± 22.5)/2
x = 8
x^3 = 512

Answer 2

The original dimensions of the box were x by x by x and, after modification, the bottom was x+.5 by
6+6, so
(x+.5)*(x+6) = 119 Go ahead and multiply it out, solve the quadratic for x, and then multiply that value by 119 to get the volume.

HTH ☺

Doug

Answer 3

Let l be the orig length
w the orig width
h the height

L = the new length
W = the new width


v=(l)(w)(h)

L = l + 0.5
W = w + 6

A =(L)(W)
119=(l+0.5)(w+6)
119=lw+0.5w+6l+3
116=lw+0.5w+6l

You don't mention if the volume is the same. There are endless solution to these since we are lacking 1 equation.

Guess: maybe the size of the paper is letter 8.5"x11".
L = 12.5" & W = 9.5"
A = (12.5)(9.5) = 118.75 sq. in.
v = (l)(w)(h)
v = (11.5)(3.5)(h)
v = 40.25h

Answer 4

You can figure out the original dimensions of the base, but I don't see enough info to find the height????