An open box with a volume of 1500cm^3 is to be constructed by taking a piece of cardboard of 20 cm by 40 cm, cutting squares of sides length x cm from each corner and folding up the sides. Show that this can be done in two different ways, and find exact dimensions of the box in each case.
I have already found the solution using the quadratic equation, can someone help me find the solution using synthetic division?
Could you explain how to do this with the quadratic equation as well?
L = 40 - 2x
w = 20 -2x
V = 1500 = L w x
1500= (40-2x)(20-2x)x = (800 -120 x + 4 x^2)x
so
4x^3 - 120 x^2 + 800 x -1500 = 0
make a reasonable guess
x = 5 for example
then the box would be 30*10 * 5 = 1500
good, height of five is a solution
so divide by (x-5)
x^3 - 120 x^2 + 800 x -1500
-----------------------------
(x-5)
= 4 x^2-100 x +300
factor that
4 (x^2- 25 x + 75) = 0
I get about 21 which is too big to take off the width and 2.89
check x = 2.89
20 - 5.78 = 14.22
40 - 5.78 = 34.22
v = 14.22*34.22*2.89 = 1406
well, better do it a bit more accurately
Why did the cardboard go to therapy?
Because it had serious issues with being cut up and folded!
Now, let's dive into solving this problem using synthetic division for a little bit of fun.
To find the dimensions of the box using synthetic division, we'll start by representing the dimensions of the cardboard as (20 - 2x) and (40 - 2x), since we're cutting squares of sides x cm from each corner.
Now, let's calculate the volume:
Volume = Length x Width x Height
1500 = x(20 - 2x)(40 - 2x)
Using synthetic division, we can simplify the equation:
First, let's rewrite the equation in descending order:
1500 = -4x³ + 120x² - 800x
Now, let's set up our synthetic division:
4 | -4 120 -800 1500
| -16 -416 1344
__________________________
-4 104 -1216 2844
After performing synthetic division, we get the simplified equation:
-4x² + 104x - 1216 + 2844/x = 0
Now, we can solve for x by factoring or using any other suitable method.
Once we find the value of x, we can substitute it back into (20 - 2x) and (40 - 2x) to get the exact dimensions of the box.
Hope this helps!
To find the solution using synthetic division, we can follow these steps:
Step 1: Let's denote the length of the side of the squares cut from each corner as 'x'. Then the length of the resulting box will be (40 - 2x) cm, and the width will be (20 - 2x) cm.
Step 2: To find the height of the box, we need to consider the volume. The volume of a rectangular box is given by the formula V = length × width × height.
In this case, the volume of the box is given as 1500 cm^3. So, we can write the equation as follows:
(40 - 2x) × (20 - 2x) × x = 1500
Step 3: Simplify the equation:
(40 - 2x) × (20 - 2x) × x - 1500 = 0
Step 4: Expand the equation:
x(40 × 20 - 40 × 2x - 20 × 2x + 4x^2) - 1500 = 0
Simplifying further:
(800 - 80x - 40x + 4x^2)x - 1500 = 0
(800 - 120x + 4x^2)x - 1500 = 0
Step 5: Use synthetic division to find the factors of the equation:
Let's assume the possible factor is 'a'. We divide the polynomial (800 - 120x + 4x^2)x - 1500 by (x - a). We are looking for the values of 'a' that will make the equation equal to zero, indicating that they are potential solutions.
Using synthetic division, we get the following results:
4 | 800 - 120 - 1500
|______________
-2000
0
The remainder is zero, indicating that 4 is a factor.
So, (x - 4) is a factor of the equation.
Step 6: Use the factor we found to write the equation as:
(x - 4)(ax + b) = 0
Now, we need to find the values of 'a' and 'b'. We can do this by expanding the equation and comparing the coefficients of the like terms with the original equation.
(a = 4, b = -200)
So, the equation can be written as:
(x - 4)(4x - 200) = 0
Step 7: Equate each factor to zero to find the possible values of 'x':
Case 1: x - 4 = 0
x = 4
Case 2: 4x - 200 = 0
4x = 200
x = 50
So, we have found two possible values for 'x': 4 and 50.
Step 8: Substitute these values of 'x' into the expressions for the length and width to find the dimensions of the box in each case:
Case 1: When x = 4
Length = 40 - 2x = 40 - 2(4) = 40 - 8 = 32cm
Width = 20 - 2x = 20 - 2(4) = 20 - 8 = 12cm
Hence, one solution is a box with dimensions 32cm × 12cm × 4cm.
Case 2: When x = 50
Length = 40 - 2x = 40 - 2(50) = 40 - 100 = -60cm (not possible since length cannot be negative)
Width = 20 - 2x = 20 - 2(50) = 20 - 100 = -80cm (not possible since width cannot be negative)
Therefore, the second case is not feasible.
Hence, the solution using synthetic division only gives one possible solution, which is a box with dimensions 32cm × 12cm × 4cm.
To find the solution using synthetic division, we can follow these steps:
Step 1: Let's assume that the original dimensions of the piece of cardboard are 20 cm by 40 cm.
Step 2: Let's assume that we cut squares of side length x cm from each corner. So, the length of the resulting open box would be (40 - 2x) cm and the width would be (20 - 2x) cm.
Step 3: To find the height of the box, we need to consider the volume. The volume of a rectangular box is given by the formula: volume = length × width × height.
In this case, the volume is given as 1500 cm^3. So, we have the equation: (40 - 2x)(20 - 2x) × x = 1500.
Step 4: Rearrange the equation to have it in the standard quadratic form: 4x^3 - 120x + 800x - 1500 = 0.
Step 5: Applying synthetic division, we set up the synthetic division table with the coefficients of the polynomial:
| 4 | 800 | -1500
____________
0
Step 6: Test different values for x using synthetic division until we find a root that results in a remainder of 0. In this case, we can try division by x = 5.
| 4 | 800 | -1500
____________
5 | 4 | 820 | 1100
| 20 | 410 | 3500
____________
4 | 840 | 3510
From this synthetic division, we can see that x = 5 is a root of the polynomial, as the remainder is 0.
Step 7: Now, we have a quadratic equation left: 4x^2 + 840x + 3510 = 0.
Step 8: To find the roots of this equation, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).
Using the values a = 4, b = 840, and c = 3510, we can plug them into the formula and solve for the roots of the equation.
By solving the quadratic equation, we find that x ≈ -104.74 or x ≈ -20.75.
Since negative values for x are not meaningful in this context, we discard the negative root.
Therefore, the only valid solution obtained using synthetic division is approximately x ≈ 20.75 cm.
Step 9: Plug this value back into the original expressions for the length and width of the box to find the exact dimensions.
Length = 40 - 2x ≈ 40 - 2(20.75) ≈ 40 - 41.5 ≈ -1.5 cm (discard)
Width = 20 - 2x ≈ 20 - 2(20.75) ≈ 20 - 41.5 ≈ -21.5 cm (discard)
Since the dimensions are negative, we can conclude that there is no valid solution using synthetic division in this case.
Therefore, there is only one valid solution to the problem, which was found using the quadratic equation method.