Question

An open rectangular box is to be constructed by cutting square corners out of a 16​- by 16​- inch piece of cardboard and folding up the flaps. A box formed this way is shown on the right. Find the value of x for which the volume of the box will be as large as possible.

Answer 1

the value of x for which the volume of the box will be as large as possible is approximately 2.67 inches.

Explanation:

Let x be the length of the side of the squares that are cut out of the corners of the cardboard. Then the dimensions of the base of the rectangular box will be:

Length = 16 - 2x (since two squares of side x are cut out of each end of the 16-inch length)

Width = 16 - 2x (since two squares of side x are cut out of each end of the 16-inch width)

The height of the box will be x, since that is the height of the squares that were cut out.

The volume of the box can be expressed as:

V = Length × Width × Height

V = (16 - 2x) × (16 - 2x) × x

V = 4x^3 - 64x^2 + 256x

To find the value of x that maximizes the volume of the box, we can take the derivative of V with respect to x and set it equal to zero:

dV/dx = 12x^2 - 128x + 256 = 0

We can solve this quadratic equation for x using the quadratic formula:

x = [128 ± sqrt(128^2 - 4 × 12 × 256)] / (2 × 12)

x = [128 ± sqrt(16384)] / 24

x = [128 ± 128] / 24

x = 10.67 or x = 2.67

Since x represents the length of a side of a square, it must be non-negative. Therefore, the only valid solution is x = 2.67 inches.

So, the value of x for which the volume of the box will be as large as possible is approximately 2.67 inches.

Answer 2

Answer: Let x be the side length of the square corners that are cut out of the cardboard. Then the dimensions of the base of the box (after the corners have been cut out and the flaps folded up) are 16-2x by 16-2x, and the height of the box is x.

The volume V of the box is given by:

V = (16-2x)(16-2x)(x)

Expanding this expression gives:

V = 4x^3 - 64x^2 + 256x

To find the value of x that maximizes V, we can take the derivative of V with respect to x and set it equal to zero:

dV/dx = 12x^2 - 128x + 256 = 0

Dividing both sides by 4 gives:

3x^2 - 32x + 64 = 0

This quadratic equation can be factored as:

(3x - 16)(x - 4) = 0

So the solutions are x = 16/3 and x = 4.

Since x must be less than half of 16 (the side length of the cardboard), we can reject x = 16/3. Therefore, the value of x that maximizes the volume of the box is x = 4 inches.

Explanation: