You are given a 60 inch by 30 inch piece of cardboad and asked to make a six-sided box. If the cardboard is cut along the lines shown, what will the dimensions of the box with maximum volume?
It would help if you would describe how the lines are drawn.
it is a rectangle with a shaded square in the top left and bottom left corners.....there is a shaded rectangle in the top right corner and bottom right corner....there is a shaded square to the left of the rectangle in the top right corner and the bottom right corner....in between the two shaded squares is an unshaded rectangle, both top and bottom....there are 2 horizonal lines that split the rectangle in 3 parts and there are 4 vertical lines that split the rectangle into four parts.
If you name the sides of the squares as of length x, then the horizontal rectangles are 30-x long, and the vertical ones are 30-2x long.
The total volume is therefore
A(x)=x(30-x)(30-2x)
which is a function of a single variable x.
Find the derivative of A(x) as A'(X) and equate A'(x)=0 to find the maximum area.
There will be two roots, x1 and x2. To distinguish the maximum from the minimum, we calculate A"(x) (second derivative of A(x)).
If A"(x)>0, it is a minimum.
If A"(x)<0, it is a maximum.
You will find that the smaller root is the maximum and equal to about 6.34, and the corresponding maximum volume is about 2598 in³.
To find the dimensions of the box with maximum volume, we need to determine the dimensions of the base and the height that will yield the largest possible volume within the given cardboard.
Let's start by visualizing the cardboard and labeling the dimensions:
--------------- 60 inches ------------
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30 inches |
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--------------- 60 inches ------------
To form a six-sided box with the given cardboard, we need to cut out squares from each corner and fold up the remaining sides. Let's decide on the side length of these squares, which we will refer to as "x".
Cutting out squares with side length x will reduce the length and width of the cardboard by 2x at each corner:
--------------- 60 inches ------------
| | | |
| | | |
| | | |
| x | x | x |
| | | |
30 inches --------------- 60 inches
This will leave us with the base dimensions of (60 - 2x) inches and (30 - 2x) inches, respectively.
To form a six-sided box, we need to fold up the remaining sides, effectively creating a solid box with a rectangular base. The height of the box will be equal to the side length x:
x
--------------- 60 inches ------------
| | | |
| | | |
x | | |
| x | | x |
| | | |
30 inches --------------- 60 inches
Now, we can calculate the volume of the box in terms of x:
Volume = length × width × height
Volume = (60 - 2x) × (30 - 2x) × x
To find the maximum volume, we need to find the value of x that maximizes the volume formula.
Let's find the derivative of the volume formula with respect to x, and set it equal to zero to find the critical points:
V'(x) = 0
Then, we can solve the resulting equation to find the value(s) of x that yield the maximum volume.
Differentiating the volume formula V(x) with respect to x:
V'(x) = (60 - 2x)(30 - 2x)x' + (30 - 2x)(60 - 2x)x'
= (60 - 2x)(30 - 2x) - (30 - 2x)(60 - 2x)
= 4x^3 - 180x^2 + 1800x
Setting V'(x) equal to zero:
4x^3 - 180x^2 + 1800x = 0
Now, we can solve this equation to find the value(s) of x.
Once we have the value of x, we can substitute it back into the volume formula to find the corresponding dimensions of the box with maximum volume.