find the original width

Question

A rectangular piece of metal is 30in 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 1050 incubed3​,

what were the original dimensions of the piece of​ metal?

what is the original width

Answer 1

1050= 6 x L-12 x W-12

(L-12)(W-12) = 1050/6 = 175

L= W+30

W= L-30

(W+30-12)(W-12) = 175

W^2 +6W = 175+ 216= 391

W^2 + 6W + 9 = 391 + 9 = 400

(W+3)^2 = 20^2

W+3 = + or - 20

W = -3 + 20 = 17

L=17+30 = 47

subtract 12 from each to get the original length and width of the rectangular metal

17-12 = 5 inches

47-12 = 35 inches

5x25x6 = 1050 cubic inches