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 -> 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 1027384: 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 josgarithmetic(39838) About Me  (Show Source):
You can put this solution on YOUR website!
w and L, the dimensions.
L=w+10, and w, again the same dimensions, using the given description.

Folding the flaps, the base of the box is (L-2*2)(w-2*2) inch^2 and the height is 2 inches.

This means the volume is 2%28L-4%29%28w-4%29=528.
Substitute for L, simplify the equation, and solve for w.

2%28w%2B10-4%29%28w-4%29=528
%28w%2B6%29%28w-4%29=528%2F2
w%5E2%2B4w-24=264
w%5E2%2B4w-24-264=0
highlight_green%28w%5E2%2B4w-288=0%29-----------Continue from here.....