A sheet metal fabricator makes boxes from sheet of steel. Each sheet of steel has an area of 1500cm2. 6cm squares are cut from each corner and the edges are bent up to form a box with a volume of 3456cm3.What are the dimensions of each box?
I was making my two equations and then using the substitution method but my equations aren't working out.
What were your equations, and what are your steps?
L × W × H = 3456cm3
L × W = 1500cm2
L= 1500cm2/W
The i plugged my second equation into my first one .
1500cm3/W × WH= 3456cm3
H= 2.304cm
I did this so far, I'm not sure if it's right and Im confused what to do next
Ok, let's go with your definitions.
original length --- l
original width ---- w
so lw = 1500
w = 1500/l
new length = l-12
new width = w-12 , you are cutting 6 cm off on each end
new height = 6
volume = 6(l-12)(w-12)= 3456
(l-12)(w-12) = 576
lw - 12l - 12w + 144 = 576
1500 - 12(l+w) + 144 = 576
1068 = 12(l+w)
l+w = 89
l + 1500/l = 89
l^2 - 89l + 1500 = 0
using the formula,
l = appr 66.4 or l = 22.6
then w = 22.6, or w = 66.4
I have symmetric solutions, so the boxes are
22.6-12 cm by 66.4-12 cm by 6 cm
or
10.6 by 54.4 by 6 cm
area = 22.6(66.4) = 1500.6 , not bad
volume = 6(22.6-12)(66.4-12) = 3459.84 , acceptable
How did you get 12. For example the equation of your new length = l-12 and w-12
You are cutting off squares of 6 cm from each end.
So the length is 6 cm less on one end, and 6 cm off the other end. Or, 12 less from both the length and the width.
Make a sketch of a rectangle the corners cut off, and you will see .
To solve this problem, let's break it down step by step:
1. Let's start by identifying the dimensions of the box. We'll assume that the original sheet of steel has a length (L) and width (W).
2. We know that 6cm squares are cut from each corner, so the resulting box will have dimensions of (L - 2*6) and (W - 2*6). This is because two 6cm squares are cut from each length and width of the original sheet.
3. Now, we have the volume of the box, which is given as 3456cm³. The formula for volume of a rectangular box is V = length * width * height. Since the height of the box is not given explicitly, we can assume it is the length of the 6cm square that was cut out (which was 6cm itself).
4. Substituting the provided values into the volume formula, we can write the equation: 3456 = (L - 12) * (W - 12) * 6
5. Next, we can use the information that the area of the original sheet is 1500cm². The formula for the area of a rectangle is A = length * width. So, we can also write the equation: 1500 = L * W
6. Now we have two equations:
- Equation 1: 3456 = (L - 12) * (W - 12) * 6
- Equation 2: 1500 = L * W
7. We can now solve this system of equations using the substitution or elimination method. You mentioned that you tried the substitution method but weren't successful. Let me show you the elimination method:
- Multiply Equation 1 by 1500, so it becomes: 20736 = (L - 12) * (W - 12) * 9000
- Expand this equation: 20736 = 9000*L*W - 32400L - 32400W + 388800
- Rearrange terms to have everything on one side: 9000*L*W - 32400L - 32400W + 366064 = 0
- Now, we can substitute Equation 2 (1500 = L * W) into the above equation:
9000 * 1500 - 32400L - 32400W + 366064 = 0
13500000 - 32400L - 32400W + 366064 = 0
- Simplify the equation further: -32400L - 32400W = -13866064
8. At this point, we have a new equation: -32400L - 32400W = -13866064
9. Now we can solve this equation to find the values of L and W, which will give us the dimensions of the box.
The steps above outline the process to solve the problem. However, I can provide the solutions to you directly if you'd like.