A carpenter wants to make an open-topped box out of a rectangular sheet of tin 24 inches wide and 45 inches long. The carpenter plans to cut congruent squares out of each corner of the sheet and then bend the edges of the sheet upward to form the sides of the box. If the box is to have greatest possible volume, what should its dimensions be?
To find the dimensions of the box with the greatest possible volume, we need to maximize the volume of the box by choosing the size of the cut-out squares.
Let's assume that the length of each cut-out square is x inches.
To form the open-topped box, the width of the box would be 24 - 2x (as two squares are cut out from the width), and the length of the box would be 45 - 2x (as two squares are cut out from the length).
The height of the box would be the size of the cut-out square, which is x inches.
The volume of the box can be calculated using the formula: Volume = Length * Width * Height
Therefore, the volume of the box would be:
V(x) = (45 - 2x) * (24 - 2x) * x
To find the value of x that maximizes the volume, we can take the derivative of V(x) with respect to x and set it equal to zero.
Let's find the derivative:
V'(x) = (45 - 2x) * (24 - 2x) + x * (-2) * (24 - 2x) + x * (-2) * (45 - 2x)
Simplifying the expression:
V'(x) = 1080 - 2x^2 - 90x
Setting V'(x) equal to zero:
1080 - 2x^2 - 90x = 0
Rearranging the equation:
2x^2 + 90x - 1080 = 0
Dividing both sides by 2:
x^2 + 45x - 540 = 0
Factoring the equation:
(x + 60)(x - 9) = 0
Solving for x:
x + 60 = 0 => x = -60 (discard as length cannot be negative)
x - 9 = 0 => x = 9
Since x represents the length of the cut-out square, we can ignore the negative value and choose x = 9 inches.
Therefore, the dimensions of the box with the greatest possible volume would be:
Width = 24 - 2x = 24 - 2(9) = 6 inches
Length = 45 - 2x = 45 - 2(9) = 27 inches
Height = x = 9 inches
So, the box should have dimensions 6 inches by 27 inches by 9 inches to have the greatest possible volume.
To find the dimensions of the box with the greatest possible volume, we need to determine the size of the congruent squares to be cut out from each corner of the rectangular tin sheet.
Let's assume the length of each side of the square to be cut out is 'x' inches.
First, let's determine the length and width of the base of the box after the squares are cut out. When the squares are cut out from each corner, the length of the base will be reduced by '2x' inches (since both ends of the sheet contribute to the length) and the width will be reduced by '2x' inches (since both sides of the sheet contribute to the width).
So, the dimensions of the base of the box will be (45 - 2x) inches and (24 - 2x) inches.
Now, the height of the box will be the length of the squares cut out, which is 'x' inches.
Therefore, the volume of the box can be calculated as the product of the length, width, and height:
Volume = (45 - 2x) * (24 - 2x) * x
To find the dimensions that result in the greatest possible volume, we need to maximize this volume equation.
We can do this by finding the value of 'x' that maximizes the volume. One approach is to differentiate the volume equation with respect to 'x', set the derivative equal to zero, and solve for 'x'. However, since we are using Explain Bot and limited in mathematical operations, we can analyze the expression and try out different values for 'x' to see which yields the maximum volume.
Let's try a few values to see which one gives the highest volume:
1. Let's assume 'x' = 1 inch:
Volume = (45 - 2 * 1) * (24 - 2 * 1) * 1
Volume = 43 * 22 * 1 = 946
2. Let's assume 'x' = 2 inches:
Volume = (45 - 2 * 2) * (24 - 2 * 2) * 2
Volume = 41 * 20 * 2 = 1640
3. Let's assume 'x' = 3 inches:
Volume = (45 - 2 * 3) * (24 - 2 * 3) * 3
Volume = 39 * 18 * 3 = 2106
Based on these trials, it seems that as 'x' increases, the volume is also increasing. However, we need to consider the constraint that 'x' should be smaller than half the dimensions of the tin sheet (since squares should be able to fit into each corner).
Considering the given dimensions of the tin sheet (24 inches wide and 45 inches long), the maximum value we can assign to 'x' is 11 inches (half of the width or length rounded down).
Let's try using 'x' = 11 inches:
Volume = (45 - 2 * 11) * (24 - 2 * 11) * 11
Volume = 23 * 2 * 11 = 506
Comparing the volumes, it seems 'x' = 3 inches results in the greatest volume of 2106 cubic inches.
Therefore, the dimensions of the box with the greatest possible volume would be:
Length: 45 - 2 * 3 = 45 - 6 = 39 inches
Width: 24 - 2 * 3 = 24 - 6 = 18 inches
Height: 3 inches
Thus, the box should have dimensions of 39 inches by 18 inches by 3 inches to achieve the greatest possible volume.
If the cuts are length x, then the volume is
v = (24-2x)(45-2x)(x)
= 4x^3 - 138x^2 + 1080x
dv/dx = 12x^2 - 276x + 1080
= 12(x^2-23x+90)
= 12(x-18)(x-5)
You can take it from there ...