A box with a square base and an open top is to have a volume of 68in^3 . Neglect the thickness of the material used to make the box, and find the dimensions of the box that would minimize the amount of material used.

The width and length of the box are each ___ in.
The height of the box is ___ in.

6x9x9

To minimize the amount of material used, we need to consider the surface area of the box.

Let's assume the width and length of the box are both "x" inches, and the height of the box is "h" inches.

The volume of the box is given by:
Volume = length * width * height

Therefore, we have the equation:
68 = x * x * h

To minimize the amount of material used, we need to minimize the surface area of the box.

The surface area of the box is given by:
Surface Area = 2lw + lh + wh

Since the box has an open top, the surface area is reduced to:
Surface Area = 2lw + lh

We want to minimize the surface area while keeping the volume constant.

Substituting the volume equation into the surface area equation, we have:
Surface Area = 2x^2 + 68/x

To find the minimum surface area, we take the derivative of the surface area equation with respect to x and set it equal to zero:
d(Surface Area)/dx = 4x - 68/x^2 = 0

Simplifying, we get:
4x - 68/x^2 = 0

Multiplying through by x^2, we have:
4x^3 - 68 = 0

Rearranging, we get:
4x^3 = 68

Dividing through by 4, we have:
x^3 = 17

Taking the cube root of both sides, we get:
x = ∛17

Therefore, the width and length of the box are both approximately ∛17 inches.

To find the height of the box, we substitute the width and length value back into the volume equation:
68 = (∛17)^2 * h

Simplifying, we have:
68 = 17h

Dividing both sides by 17, we get:
h = 68/17 = 4

Therefore, the height of the box is 4 inches.

The width and length of the box are each approximately ∛17 inches, and the height of the box is 4 inches.

To minimize the amount of material used, we need to minimize the surface area of the box.

Let's denote the width and length of the square base as x, and the height of the box as h.

The volume of the box is given as 68in^3, which means we have the equation:

Volume = width * length * height
68 = x * x * h
68 = x^2 * h

To minimize the surface area, we need to express it in terms of a single variable. The surface area of the box consists of the area of four sides (which are all equal since the base is a square) and the area of the bottom:

Surface Area = 4 * (width * height) + (width * length)
Surface Area = 4 * (x * h) + (x * x)
Surface Area = 4xh + x^2

Now, let's express the surface area in terms of just one variable. Since we have the equation 68 = x^2 * h, we can solve for h:

h = 68 / x^2

Substituting this value back into the surface area expression:

Surface Area = 4x(68 / x^2) + x^2
Surface Area = 272 / x + x^2

To minimize the surface area, we can take the derivative of the surface area equation with respect to x, set it equal to zero, and solve for x:

d(Surface Area) / dx = -272 / x^2 + 2x

Setting this equal to zero:

-272 / x^2 + 2x = 0

To solve this equation, we can multiply through by x^2 to eliminate the fractions:

-272 + 2x^3 = 0

Rearranging the equation:

2x^3 = 272

Dividing through by 2:

x^3 = 136

Taking the cube root of both sides:

x = ∛136 ≈ 5.77

Now, we can substitute this value of x back into the equation h = 68 / x^2 to find the value of h:

h = 68 / (5.77)^2 ≈ 2.42

Therefore, the width and length of the box are each approximately 5.77 inches, and the height of the box is approximately 2.42 inches.