Asked by diana
A PIECE OF SHEET METAL IS 2.6 TIMES AS LONGS AS IT IS WIDE.
IT IS TO BE MADE INTO A BOX WITH AN OPEN TOP BY CUTTING 3-INCH SQUARES FROM EACH CORNER AND FOLDING UP THE SIDES.
IF THE VOLUME OF THE BOX MUST BE BETWEEN 600 AND 800 CUBIC INCHES, WHAT VALUES OF X WILL PRODUCE THIS RANGE OF VOLUMES?
IT IS TO BE MADE INTO A BOX WITH AN OPEN TOP BY CUTTING 3-INCH SQUARES FROM EACH CORNER AND FOLDING UP THE SIDES.
IF THE VOLUME OF THE BOX MUST BE BETWEEN 600 AND 800 CUBIC INCHES, WHAT VALUES OF X WILL PRODUCE THIS RANGE OF VOLUMES?
Answers
Answered by
Damon
width of sheet = x ??? I will assume
b, breadth of box = x - 6
L = length of box = 2.6 x - 6
height = 3
600 ≤ 3(x-6)(2.6 x - 6) ≤ 800
600 ≤ 7.8 x^2 -64.8 x + 108 ≤ 800
492 ≤ 7.8 x^2 - 64.8 x ≤ 692
solve those quadratics (ignore negative solutions)
x = 13.1
x = 14.4
b, breadth of box = x - 6
L = length of box = 2.6 x - 6
height = 3
600 ≤ 3(x-6)(2.6 x - 6) ≤ 800
600 ≤ 7.8 x^2 -64.8 x + 108 ≤ 800
492 ≤ 7.8 x^2 - 64.8 x ≤ 692
solve those quadratics (ignore negative solutions)
x = 13.1
x = 14.4
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