What is the original width in inches?

Question

A rectangular piece of metal is 30 in longer than it is wide. Squares with sides

6 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 2400

in^3​, what were the original dimensions of the piece of​ metal

Answer 1

L = 30+w inches

corners cut out, open box created with volume = 2400 cubic inches

what were dimensions of the original sheet, L and w?? Length and width? 52 inches by 22 inches

6(52-12)(22-12) = 6(40)(10) = 6(400) = 2400 in^3

you could have chosen endless other squares to cut out other than 6 inch squares

x(L-2x)(w-2x) = 2400

x(w+30-2x)(w-2x) = 2400

(L-2x)(w-2x) = Lw - 4x^2

(w+30-2x)(W-2x = (w+30)w -4x^2

down to 2 equations 2 unknowns

expand and solve by substitution elimination

w^2 +30x -2xw -2xw +60x +4x^2 = w^2 +30w -4x^2

30x -4xw +30w = 0

it's possible there is no solution as the cubic and quadratic may never intersect

L=72, w=42, x=1 is close but volume then = 2800, not 2400

close but no cigar

or there is an infinite number of values of L,w and x that will make V=2400

L=52, w=22, x =6

then smaller rectangle is 10 by 40 by 6 = 2400 cubic inches