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?
The diagram looks as follows:
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.
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Suppose you want to make an open-topped box out of a 3 \times 6 index card by cutting a square out of each corner and then folding up the edges. How large a square should you cut out of each corner in order to maximize the volume of the resulting box?
To find the dimensions of the box with maximum volume, we need to determine how to best use the given cardboard to form a six-sided box.
Looking at the diagram, we can see that the shaded square in the top left and bottom left corners will form the base of the box. Let's call the side length of each square "x".
The shaded rectangle in the top right and bottom right corners will be the height of the box. Let's call the width of this rectangle "y".
The unshaded rectangle between the two shaded squares will be the length of the box. Let's call the length of this rectangle "z".
With this information, we can start determining the dimensions of the box.
From the diagram, we can see that the length of the piece of cardboard is 60 inches and the width is 30 inches. Since we will be using the shaded squares as the base, we can use them to find the value of "x".
The length of each shaded square is given by (30 - 2x), and the width is (30 - x).
Since there are two squares, we can write the equation:
2(30 - 2x)(30 - x) = area of the base
To find the height, we can look at the shaded rectangle in the top right and bottom right corners. The width of this rectangle is already given as "y".
The length of this rectangle is given by (60 - 2y - z), because the total length of the cardboard is 60 inches, and we have already accounted for the shaded squares.
Finally, the length of the box is given by the unshaded rectangle in between the two shaded squares, which is just "z".
The volume of the box is given by multiplying the dimensions together:
Volume = (30 - 2x)(30 - x)(60 - 2y - z)
To find the dimensions of the box with maximum volume, we need to maximize this volume equation. This can be done by taking the derivative of the volume equation with respect to the variables "x", "y", and "z", and setting them equal to zero.
Solving these equations will give us the values of "x", "y", and "z" that result in the maximum volume for the box.