a square sheet of tin 30cm on a side is to be used to make an open - top box by cutting a small square of tin from each corner and bending up the sides. how large should be the square cut from each corner to make the box's volume as large as possible
Well, well, well, looks like we've got some tin trouble here! To maximize the volume of this open-top box, we need some clever thinking. Let's embark on this amusing mathematical journey, shall we?
First off, we have a square sheet of tin with side length 30cm. To make an open-top box, we need to cut square corners from that sheet. The size of these squares will determine the height of the box.
Let's assume we cut a square of size 'x' from each corner. Thus, the dimensions of the resulting box will be (30 - 2x) cm by (30 - 2x) cm, with a height of 'x' cm.
So, the volume of the box can be calculated as V = (30 - 2x) * (30 - 2x) * x.
To maximize the volume, we can go down the path of differentiation! But fret not, for this will be a mild mathematics jest.
Differentiating V with respect to x, we get:
dV/dx = 4x^2 - 120x + 900.
To find the maximum (or minimum) point, we set this derivative equal to zero:
4x^2 - 120x + 900 = 0.
Solving this quadratic equation, we find two solutions, x = 7.5 and x = 15.
Now, my comedic comrade, keep in mind that x represents the size of the square we cut from each corner. Since we're dealing with a physical square sheet of tin, the cut-out squares cannot have a negative size, you see. Therefore, we'll discard the x = 15 solution.
So, after all this numerically comedic calculation, we find that cutting squares with a size of 7.5 cm from each corner will result in the box having the largest possible volume.
So grab your tin snips and get to cutting, my friend, because with this information, your open-top box will be at its volumetric zenith! Enjoy the process!
To find the dimensions of the square cut from each corner that will maximize the volume of the box, we need to follow these steps:
Step 1: Visualize the problem
- Imagine a square sheet of tin with side length 30 cm.
Step 2: Define the variables
- Let x be the side length of the square cut from each corner.
Step 3: Determine the dimensions of the box
- When we cut squares of size x from each corner and fold up the sides, the resulting box will have a length of (30 - 2x) cm, a width of (30 - 2x) cm, and a height of x cm.
Step 4: Calculate the volume
- The volume, V, of the box can be calculated as:
V = length × width × height
V = (30 - 2x) × (30 - 2x) × x
Step 5: Simplify the volume equation
- Expand the equation:
V = (900 - 60x - 60x + 4x^2) × x
V = (900 - 120x + 4x^2) × x
V = 4x^3 - 120x^2 + 900x
Step 6: Maximize the volume
- To maximize the volume, we need to find the value of x that gives the maximum volume.
- We can do this by taking the derivative of the volume equation with respect to x, setting it equal to zero, and solving for x.
Step 7: Calculate the derivative
- Differentiate the volume equation:
V' = 12x^2 - 240x + 900
Step 8: Set the derivative equal to zero
- Set V' = 0 and solve for x:
12x^2 - 240x + 900 = 0
Step 9: Solve for x
- Solve the quadratic equation using factoring, completing the square, or quadratic formula to find the critical points.
Step 10: Evaluate the critical points
- Evaluate the critical points to determine which one corresponds to a maximum volume.
Step 11: Calculate the maximum volume
- Substitute the value of x that maximizes the volume into the volume equation:
V = 4x^3 - 120x^2 + 900x
In this way, you can determine the value of x that will maximize the volume of the box.
To determine the size of the small square cut from each corner to maximize the volume of the open-top box, we need to follow several steps:
1. Visualize the problem: Draw a square with side length 30cm and label it as the original tin sheet.
2. Determine the side length of the square cut from each corner: Let's assume that we cut a small square with side length x cm from each corner. This will leave us with a larger square in the center.
3. Calculate the dimensions of the open-top box: After cutting the squares and bending the sides, the box will have a length of 30cm - 2x (since we cut x cm from both ends), a width of 30cm - 2x (for the same reason), and a height of x cm.
4. Write an equation for the volume of the box: The volume (V) of a rectangular box is given by multiplying its length (L), width (W), and height (H): V = L x W x H.
5. Substitute the dimensions into the volume equation: V = (30cm - 2x) x (30cm - 2x) x x.
6. Simplify the equation: Multiply out the terms and express the equation in terms of x only: V = x(30cm - 2x)(30cm - 2x).
7. Maximize the volume: To find the maximum volume, we need to find the value of x that yields the highest result. This can be done using calculus by finding the derivative of the volume equation with respect to x, setting it equal to zero, and solving for x. However, since x represents a length, the maximum value occurs at the midpoint, so we can bypass calculus and simply set x = (30cm - 2x)/2.
8. Solve for x: Rearrange the equation to isolate x: 2x = 30cm - 2x.
9. Simplify and solve for x: Add 2x to both sides of the equation: 4x = 30cm. Divide both sides by 4: x = 7.5cm.
10. Calculate the dimensions of the box: Substitute x = 7.5cm into the dimensions equation to find the final dimensions of the box. The length and width will be 30cm - 2(7.5cm) = 15cm, and the height will be 7.5cm.
Therefore, to maximize the volume of the open-top box, each corner should be cut with a square measuring 7.5cm on each side.
let the size of the cut-out square be x cm by x cm
then after bending,
the base of the box will be 30-2x by 30-2x and x cm high
Volume = x(30-2x)^2
= x(900 - 120x + 4x^2)
= 4x^3 - 120x^2 + 900x
d(Volume)/dx = 12x^2 - 240x + 900
= 0 for a max/min of Volume
12x^2 - 240x + 900 = 0
x^2 - 20x + 75 = 0
(x-5)(x-15) = 0
x = 5 or x = 15 , (clearly x = 15 yields a minimum volume of 0)
So the size of the cut should be squares of 5 cm by 5 cm.