An open-top box is to be constructed from a 6 foot by 8 foot rectangular cardboard by cutting out equal squares at each corner and the folding up the flaps. Let x denote the length of each side of the square to be cut out.

Question

An open-top box is to be constructed from a 6 foot by 8 foot rectangular cardboard by cutting out equal squares at each corner and the folding up the flaps. Let x denote the length of each side of the square to be cut out.

a)

Answer 1

To find the value of x that will maximize the volume of the box, we need to set up an equation for the volume and then maximize it using calculus.

Let's start by drawing a diagram of the box. The original rectangular cardboard has dimensions 6 feet by 8 feet.

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| Box |
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Now, when we cut out the squares at each corner and fold up the flaps, the remaining part forms the base and sides of the box. Let's label the dimensions of the box as follows:

x 6-2x x
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| Box | 8-2x
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|__________________________________|

The length and width of the base are given by 6-2x and 8-2x, respectively, and the height will be x.

The volume of the box is given by the product of its length, width, and height:

Volume = (6-2x)(8-2x)(x)

To maximize the volume, we need to find the value of x that maximizes this function.

To do this, we can take the derivative of the volume function with respect to x and set it equal to zero. Then, solve the resulting equation to find the critical point(s) which will give us the values of x that maximize the volume.

Let's differentiate the volume function:

d/dx [(6-2x)(8-2x)(x)]

To simplify the expression, we can use the product rule:

d/dx [(6-2x)(8-2x)(x)] = (8-2x)(x) + (6-2x)(x) + (6-2x)(8-2x)

Expanding the expression further:

= 8x - 2x^2 + 6x - 2x^2 + 48x - 12x - 4x^2 + 24x

= -8x^2 + 86x

Setting this expression equal to zero:

-8x^2 + 86x = 0

Now, we can factor out x:

x(-8x + 86) = 0

So, we have two possible solutions:

x = 0
-8x + 86 = 0

Solving the second equation:

-8x + 86 = 0

Adding 8x to both sides:

86 = 8x

Dividing by 8:

x = 10.75

Since the length of each side of the square cannot be zero in this context, we disregard x = 0.

So, our final answer is x = 10.75 feet.

Therefore, to maximize the volume of the box, squares with sides of length 10.75 feet should be cut out at each corner of the 6x8 rectangular cardboard.