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

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: A rectangular piece of metal is 10 in longer than it is wide. Squares with sides 2 in long are cut from the four corners and the flaps are folded upward to form an open box. If the      Log On


   



Question 1027389: A rectangular piece of metal is 10 in longer than it is wide. Squares with sides 2 in long are cut from the four corners and the flaps are folded upward to form an open box. If the volume of the box is 528in^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!
let width of metal plate be x
length = x+10
after cutting the corners
width = (x-4)
and
Length = (x+10)-4= x+6
The height will be 2 inches
Volume = L * W * H
(x-4)(x+6)*2=528
(x-4)(x+6) = 264
x^2+2x-24=264
x^2+2x =288
x^2+2x+1= 288
(x+1)^2=289
take the square root
(x+1) = +/- 17
x= 16 or -18
ignore negative
width = 16in and length = 26in