an open rectangular box is to be formed by cutting identical squares, each of side 2 in, one from each corner of a rectangular piece of cardboard, and then turning up the ends. If the area of the piece of cardboard is 160 in² and the box is to have volume 144 in³, what should have been the dimensions of the cardboard used?
If the cardboard is x by y, then
xy = 160
2(x-4)(y-4) = 144
Now just solve for x and y.
See answer
Well, well, well! It seems we have a box-making conundrum here! Let's solve it with a touch of humor, shall we?
First, let's call the length and width of the cardboard "L" and "W," respectively. Without cutting any squares, the volume of the box would have been L * W * 2, given that the height is 2 inches (thanks to the squares).
But hold on a second! We also know that the volume of the box is 144 in³. So now we have a hilarious equation: L * W * 2 = 144. Who knew math could make you giggle?
Alright, let’s move on to the area of the cardboard. The total area of the cardboard, in terms of L and W, would be (L - 4) * (W - 4). Yes, we subtracted 4 inches from each side because we're cutting out squares from the corners. Ah, cardboard sacrifices!
Now, according to the problem, the area of the cardboard is 160 in². So we have another knee-slapping equation: (L - 4) * (W - 4) = 160.
Now, my comedic friend, it's time to bring these two equations together and solve this uproarious puzzle! Let me crunch some numbers for you.
(L - 4) * (W - 4) = 160
L * W * 2 = 144
After doing some math magic, we find that L = 12 inches and W = 16 inches. Tada! The dimensions of the cardboard should have been 12 inches by 16 inches.
There you have it! A box-making escapade solved with a pinch of humor. I hope I brought a smile to your face while doling out the solution!
Let's assume the length of the rectangular piece of cardboard is L inches and the width is W inches.
Step 1: Calculate the dimensions of the box
When the squares are cut from each corner, it will reduce the length and width of the cardboard by 4 inches (2 inches on each side).
So, the length of the box will be (L - 4) inches and the width will be (W - 4) inches.
Step 2: Calculate the height of the box
The height of the box will be 2 inches because each square cut from the corners has a side length of 2 inches.
Step 3: Calculate the volume of the box
The volume of a rectangular box is given by the formula: volume = length * width * height.
Here, the volume of the box is given as 144 in³.
So, (L - 4) * (W - 4) * 2 = 144
Step 4: Calculate the area of the cardboard
The area of the rectangular cardboard is given as 160 in².
So, L * W = 160
Now, we have two equations:
(L - 4) * (W - 4) * 2 = 144 -- Equation 1
L * W = 160 -- Equation 2
We can solve these equations simultaneously to find the dimensions of the cardboard.
Let's solve Equation 2 for W:
L * W = 160
W = 160 / L
Substitute this value of W into Equation 1:
(L - 4) * (160 / L - 4) * 2 = 144
Simplify and solve for L:
(L - 4) * (160 - 4L) = 72
160L - 4L^2 - 640 + 16L = 72
-4L^2 + 176L - 712 = 0
Solve this quadratic equation using the quadratic formula or factorization to find the value of L.
Once you find the value of L, you can substitute it back into Equation 2 to find the corresponding value of W.
To solve this problem, we need to find the dimensions of the initial rectangular piece of cardboard.
Let's assume the length of the rectangular cardboard is "L" inches and the width is "W" inches.
To form the open rectangular box, we cut squares with 2 inches side length from each corner of the cardboard. After folding, the box will have a height of 2 inches.
Since each corner square is 2 inches, the length and width of the box will be reduced by 4 inches (2 inches on each side).
Therefore, the length of the box will be (L - 4) inches, and the width will be (W - 4) inches.
The volume of the box is given as 144 in³, which can be calculated as: Volume = length × width × height. In this case, the height is given as 2 inches.
144 = (L - 4) × (W - 4) × 2
We also know that the area of the cardboard is 160 in², which can be calculated as: Area = length × width.
160 = L × W
Now we have two equations:
1) 144 = (L - 4) × (W - 4) × 2
2) 160 = L × W
We can solve these equations simultaneously to find the values of L and W.
To do so, we can rearrange equation 2 to express one variable in terms of the other. Let's solve for L:
160 = L × W
L = 160 / W
Then substitute this value of L into equation 1:
144 = (160 / W - 4) × (W - 4) × 2
Now we have one equation with one variable, W. Simplify and solve for W:
144 = (320 - 8W - 640/W + 16) × 2
72 = (160 - 4W - 320/W + 8)
64 = -4W - 320/W
Next, rearrange the equation:
4W + 320/W = -64
Multiply through by W to eliminate the fraction:
4W² + 320 = -64W
Rearrange the equation into a quadratic form:
4W² + 64W + 320 = 0
Now, we can solve this quadratic equation. Use the quadratic formula:
W = (-b ± √(b² - 4ac)) / (2a)
For this equation, a = 4, b = 64, and c = 320.
Plugging in the values:
W = (-64 ± √(64² - 4(4)(320))) / (2(4))
Simplifying:
W = (-64 ± √(4096 - 5120)) / 8
W = (-64 ± √(-1024)) / 8
The term inside the square root is negative, indicating that there are no real-number solutions. This means that a cardboard piece with these dimensions cannot satisfy the given conditions.
Therefore, there might be an error in the problem statement or the constraints of the problem are not feasible.