Algebra Question

Question

A rectangular piece of metal is 25 in longer than it is wide. Squares with sides 5 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 2730^(3), what were the original dimensions of the piece of metal?

Answer 1

volume = LWH = length x width x height

L=W+25

H=5 inches

Volume = 2730 cubic inches

2730 = (W+25)W(5) = 5W^2 + 75W

5W^2 +75W - 2730 = 0

W^2 + 15W - 546 = 0

use quadratic formula, completing the square or factor

W^2 + 15W + 225/4 = 546 + 225/4

(W+15/2)^2 = 2409/4

W +15/2 = (1/2)sqr2409 =about + or - (1/2)(49.1) =+ or - 24.6; ignore the negative square root

W= 24.6-7.5 = 17.1 inches

L=25+17.1 = 42.1 inches

but add 10 inches to both, to get the original dimensions, before the corners were cut out

original W = 17.1+10 = 27.1 inches

original L = 42.1 + 10 = 52.1 inches