SOLUTION: A rectangular piece of metal is 10 in longer than it is wide. Squares with sides 2 in Long are cut from the corners, and the flaps are folded upward to form an open box. If the vol

Algebra ->  Rectangles -> SOLUTION: A rectangular piece of metal is 10 in longer than it is wide. Squares with sides 2 in Long are cut from the corners, and the flaps are folded upward to form an open box. If the vol      Log On


   

Question 782142: A rectangular piece of metal is 10 in longer than it is wide. Squares with sides 2 in Long are cut from the corners, and the flaps are folded upward to form an open box. If the volume of the box is 832 in^3 what were the original dimensions of the piece of metal ?
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
width = x
length = x+10 in
2 inch is cut from both ends
so width = x-2
and
length = x+10-2=x+8
The dimensions of the box
x-2-----width
x+8 length
2inch = height

Volume = L * W *h
(x-2)(x+8)*2=832
x^2+6x-16=832/2
x^2+6x-16=416
x^2+6x-432=0
Find the roots of the equation by quadratic formula

a= 1 b= 6 c= -432 2916

b^2-4ac= 36 - -1728
b^2-4ac= 1764 sqrt%28%091764%09%29= 42
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29 )/
x1=( -6 + 42 )/ 2
x1= 18
x2=( -6 - 42 )/ 2
x2= -24
Ignore negative value
x = 18
width = 18 inch
length = 10+8 =18 inc