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?
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
To determine the values of x that will produce a volume between 600 and 800 cubic inches, let's go through the steps:
1. Let's assign variables to the dimensions of the sheet metal:
- Length: L
- Width: W
- Height (after folding up the sides): H
2. We know that the sheet metal is 2.6 times as long as it is wide, so we can express the relationship between the length and width as:
L = 2.6W
3. To calculate the volume of the box, we need to determine the height (H). After cutting the squares from each corner and folding up the sides, the height will be equal to the length of the square cutouts (3 inches).
4. The volume of the box can be calculated using the formula: Volume = Length × Width × Height
V = (L - 2x) × (W - 2x) × 3
where x represents the length of each side of the square cutout.
5. Substitute L = 2.6W into the volume formula:
V = (2.6W - 2x) × (W - 2x) × 3
6. Expand the formula:
V = 7.8W^2 - 6.2Wx + 12x^2
7. We are given that the volume must be between 600 and 800 cubic inches, so we can set up the following inequality:
600 ≤ 7.8W^2 - 6.2Wx + 12x^2 ≤ 800
8. To solve this inequality, we can convert it into two separate inequalities:
a) 7.8W^2 - 6.2Wx + 12x^2 ≥ 600
b) 7.8W^2 - 6.2Wx + 12x^2 ≤ 800
Now we have set up the mathematical expression for the problem. To solve these inequalities, we need values for W and x. Do you have any specific values for W that we should consider?
To solve this problem, let's start by determining the dimensions of the sheet metal. Let's say the width of the sheet metal is "x" inches. Since the length is 2.6 times the width, the length of the sheet metal will be 2.6x inches.
To make a box with an open top, we need to cut 3-inch squares from each corner of the sheet metal. This will reduce the width and length of the sheet metal by 6 inches (3 inches on each side).
After cutting the squares and folding up the sides, the dimensions of the box will be as follows:
Width: x - 6 inches
Length: 2.6x - 6 inches
Height: 3 inches
To find the volume of the box, we multiply the width, length, and height:
Volume = (x - 6) * (2.6x - 6) * 3
Now, we need to find the range of values for x that will produce a volume between 600 and 800 cubic inches.
Let's set up the inequalities based on this requirement:
600 <= (x - 6) * (2.6x - 6) * 3 <= 800
We can solve this inequality to find the values of x that satisfy the volume requirement. To simplify the process, let's divide the entire inequality by 3:
200 <= (x - 6) * (2.6x - 6) <= 266.67
Now, let's solve the inequalities separately:
For the lower bound:
200 <= (x - 6) * (2.6x - 6)
Expand the equation:
200 <= 2.6x^2 - 6x - 15.6x + 36
Combine like terms:
200 <= 2.6x^2 - 21.6x + 36
Rearrange the equation to form a quadratic inequality:
2.6x^2 - 21.6x + 36 - 200 >= 0
2.6x^2 - 21.6x - 164 >= 0
Now, we can solve this quadratic inequality using factoring, quadratic formula, or graphing methods. Once we find the solution, we can write it as x >= [solution].
For the upper bound:
(x - 6) * (2.6x - 6) <= 266.67
Expand the equation:
2.6x^2 - 6x - 15.6x + 36 <= 266.67
Combine like terms:
2.6x^2 - 21.6x + 36 <= 266.67
Rearrange the equation to form a quadratic inequality:
2.6x^2 - 21.6x + 36 - 266.67 <= 0
2.6x^2 - 21.6x - 230.67 <= 0
Again, we can solve this quadratic inequality using factoring, quadratic formula, or graphing methods. Once we find the solution, we can write it as x <= [solution].
By solving the individual inequalities, we can find the range of values for x that will produce the desired range of volumes for the box.