If an open box has a square base and a volume of 106 in.3 and is constructed from a tin sheet, find the dimensions of the box, assuming a minimum amount of material is used in its construction. (Round your answers to two decimal places.)

height in
length in
width in

To find the dimensions of the box that requires a minimum amount of material, we can use calculus. Let's assume the side length of the base of the box is x inches.

Since the base is square, its area is x^2 square inches.

Given that the volume of the box is 106 cubic inches, we can set up the following equation:

x^2 * h = 106

where h represents the height of the box.

To minimize the amount of material used, we need to minimize the surface area of the box, which is the sum of the areas of the base and the four sides.

The area of the base is x^2 square inches, and the four sides are rectangles with dimensions x inches by h inches.

Therefore, the surface area, A, is given by:

A = x^2 + 4(x * h)

Now, we can substitute the value of h from the volume equation:

A = x^2 + 4(x * (106/x^2))

Simplifying further:

A = x^2 + 424/x

To find the minimum amount of material used, we need to minimize the surface area. This can be accomplished by taking the derivative of A with respect to x, setting it equal to zero, and solving for x.

dA/dx = 2x - 424/x^2 = 0

Multiplying through by x^2:

2x^3 - 424 = 0

Solving for x:

2x^3 = 424

x^3 = 212

Taking the cube root:

x = ∛212

x ≈ 6.07 inches

Now, we can substitute this value of x back into the volume equation to find the height, h:

6.07^2 * h = 106

h ≈ 106 / (6.07^2)

h ≈ 2.97 inches

Therefore, the dimensions of the box, assuming a minimum amount of material is used in its construction, are:

Height: approximately 2.97 inches
Length: approximately 6.07 inches
Width: approximately 6.07 inches

To find the dimensions of the open box, we need to consider the relationship between the volume of the box and its dimensions.

The volume of a rectangular box can be calculated by multiplying its length, width, and height. In this case, we are given that the volume is 106 in³.

Let's assume the length of the square base to be x inches.

Since the base is square, the width will also be x inches.

The height of the box, H, will be the remaining dimension.

So, the volume equation becomes:
V = x * x * H = 106 in³

Now, we can solve for H by rearranging the equation:
H = 106 / (x * x)

The goal is to minimize the amount of material used. This means we need to minimize the surface area of the box. The surface area is the sum of the areas of the base, the top, and the four sides.

The area of the base is simply x * x = x² square inches.

The top has the same area as the base, so it will also be x² square inches.

The four sides are identical rectangles with dimensions x by H, so their total area is 4 * x * H = 4 * x * (106 / (x * x)) = 424 / x square inches.

Therefore, the total surface area, A, is given by:
A = 2 * (x²) + (424 / x)

Since we want to minimize the amount of material used, we need to minimize the surface area. To do that, we can take the derivative of A with respect to x and set it equal to zero to find the critical points.

dA/dx = 4x - 424 / x²

Setting dA/dx = 0 gives:
4x - 424 / x² = 0

Let's solve this equation to find the critical points:

Multiplying both sides by x²:
4x³ - 424 = 0

Adding 424 to both sides:
4x³ = 424

Dividing both sides by 4:
x³ = 106

Taking the cube root of both sides:
x = ∛106 ≈ 4.77

Now, we can substitute this value of x back into the equation for H to find the height:

H = 106 / (x * x) = 106 / (4.77 * 4.77) ≈ 4.45

Therefore, the dimensions of the open box, assuming a minimum amount of material is used, are approximately:
Height: 4.45 inches
Length and Width: 4.77 inches

V = L*W*H = 106in^3,

Since the base is square,
W = H.
L*W*W = 106,
L*W^2 = 106,
To minimize the material,
let L = W:
W*W*W = 106,
W^3 = 106,
W = crt(106) = 4.73in.
L = 4.73in.
H = 4.73in.