A box without a lid will be constructed from 75 cm x 100 cm piece of cardboard, by cutting squares of the same size from each corner, and folding up the sides. What is the approximate volume of the largest possible box that can be constructed?
If the squares have length x, then
v = (75-2x)(100-2x)x = 4x^3 - 350x^2 + 7500x
dv/dx = 12x^2 - 700x + 7500
dv/dx = 0 at x = (175 - 25√13)/6 ≈ 14.14
So the maximum volume is about 47,380 cm^2
If you don't yet have calculus tools, then you'll need a numeric method or a graph to find the value of x needed.
Well, if we're talking about a box without a lid, it sounds like we're dealing with a bottomless box. So basically, it's more like a tray than a box. Let's call it the world's most impractical snack tray!
To find the largest possible volume, we need to maximize the height of the box. So, we'll cut the squares from each corner and fold up the sides to create a rectangular shape. In this case, the length of the box will be 75 cm - 2x, and the width will be 100 cm - 2x, where x is the size of the square we cut from each corner.
Now, to maximize the volume, we need to make the length, width, and height as large as possible. However, since the length and width are both decreasing with each cut, we want to minimize the size of the cut x in order to maximize the volume. So, we choose x to be as small as possible, which means making the square as tiny as possible.
If we make x equal to zero, we won't be cutting anything, and we can just use the original dimensions of the cardboard as the dimensions of the box. So, the approximate volume of the largest possible box would be:
Volume ≈ length x width x height
≈ (75 cm - 2x) x (100 cm - 2x) x x
Now, since x is really small, we can approximate the volume to:
Volume ≈ 75 cm x 100 cm x x
≈ 7500 cm^3 x
So, the approximate volume of the largest possible box would be 7500x cm³. But keep in mind that this is just an approximation, and the actual volume could be slightly different.
To find the largest volume of the box, we need to maximize the height as much as possible.
Let's assume that we are cutting squares with sides of length "x" from each corner.
The dimensions of the resulting box will be:
Length = 100 cm - 2x
Width = 75 cm - 2x
Height = x
To maximize the volume, we need to find the value of "x" that will result in the largest possible volume.
The volume of the box is given by:
Volume = Length x Width x Height
Volume = (100 - 2x) x (75 - 2x) x x
Now, we can differentiate the volume function with respect to "x" and set it equal to zero to find the critical points:
dV/dx = 0
2(100 - 2x)(75 - 2x) + (100 - 2x)(-2) = 0
(100 - 2x)(75 - 2x) - (100 - 2x) = 0
Solving this equation will give us the value of "x" that will maximize the volume.
Expanding the equation and simplifying, we get:
(100 - 2x)(75 - 2x) - (100 - 2x) = 0
7500 - 200x - 150x + 4x^2 - 100 + 2x = 0
4x^2 - 352x + 7400 = 0
Using the quadratic formula, we find the values of "x" that solve this equation are approximately -3.3 and 56.3. Since the length and width of the cardboard piece are only 100 cm and 75 cm, we can discard the negative value.
Therefore, the approximate value of "x" that maximizes the volume is 56.3 cm.
Now, we can substitute this value of "x" into the volume equation to find the maximum volume:
Volume = (100 - 2x) x (75 - 2x) x x
Volume = (100 - 2(56.3)) x (75 - 2(56.3)) x 56.3
Volume ≈ 183746.3 cm^3
So, the approximate volume of the largest possible box that can be constructed is 183746.3 cm^3.
To find the largest possible volume of the box, we need to determine the size of the squares that will be cut from each corner of the cardboard.
Let's assume the size of the square cut from each corner is "x". When the squares are cut and the sides are folded up, the dimensions of the resulting box will be:
Length: (100cm - 2x)
Width: (75cm - 2x)
Height: x
The volume of a rectangular box is given by the formula: Volume = Length * Width * Height.
Using the given dimensions, we can substitute into the formula:
Volume = (100cm - 2x) * (75cm - 2x) * x
To find the largest possible volume, we need to maximize this expression. To do that, we can take the derivative with respect to 'x', set it equal to zero, and solve for 'x'. This will give us the value of 'x' that maximizes the expression.
Differentiating the expression with respect to 'x', we get:
d(Volume)/dx = 3750 - 300x - 4x^2
Setting this equal to zero, we have:
3750 - 300x - 4x^2 = 0
Simplifying, we get:
4x^2 + 300x - 3750 = 0
Now we can solve this quadratic equation for 'x'. There are different ways to solve it, but let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 4, b = 300, and c = -3750. Substituting these values into the quadratic formula, we have:
x = (-300 ± √(300^2 - 4*4*(-3750))) / (2*4)
Simplifying further, we get:
x = (-300 ± √(90000 + 60000)) / 8
x = (-300 ± √150000) / 8
Now we can evaluate the two solutions:
x₁ = (-300 + √150000) / 8
x₂ = (-300 - √150000) / 8
Using a calculator, we find:
x₁ ≈ 5.86 cm
x₂ ≈ -50.86 cm
Since the size of the squares cannot be negative, we discard x₂. Therefore, the approximate size of the squares to be cut from each corner is x = 5.86 cm.
Substituting this value back into the expression for Volume, we have:
Volume = (100cm - 2*5.86cm) * (75cm - 2*5.86cm) * 5.86cm
Calculating this, we find:
Volume ≈ 28435 cm³
Hence, the approximate volume of the largest possible box that can be constructed is approximately 28435 cubic centimeters.