A container company is tasked to make an open-top rectangular box with a square base. The box must have a volume of 108cm^(3). let the length of the sides of the square base be x cm and the height h cm.

(1) what value of x will produce the minimum outer surface area?
(2) What is the minimum surface area

Volume = x^2 h = 108 = constant

Surface Area = A (x) = x^2 + 4xh
= x^2 + 4x*108/x^2
= x^2 + 432/x

For minimum area,
dA/dx = 0
2x -432/x^2 = 0
x^3 = 216
x = 6 cm (and h = 108/36 = 3)

Minimum area = 36 + (432/6) = 108 cm^2

(1) To find the value of x that will produce the minimum outer surface area, we need to find the derivative of the surface area function with respect to x, set it equal to zero, and solve for x.

The surface area of the box is given by: A = x^2 + 4xh.

First, we'll express h in terms of x using the volume of the box:

V = x^2 * h
108 = x^2 * h
h = 108/x^2

Now substitute h into the surface area equation:

A = x^2 + 4x * (108/x^2)
A = x^2 + 432/x

To find the minimum, we'll differentiate A with respect to x:

A' = 2x - 432/x^2

Setting A' equal to zero:

2x - 432/x^2 = 0
2x = 432/x^2
2x^3 = 432
x^3 = 216
x = βˆ›216
x = 6 cm

Therefore, the value of x that will produce the minimum outer surface area is 6 cm.

(2) To find the minimum surface area, substitute the value of x into the surface area equation:

A = (6 cm)^2 + 432/(6 cm)
A = 36 cm^2 + 72 cm
A = 108 cm^2

Therefore, the minimum surface area of the box is 108 cm^2.

To find the value of x that will produce the minimum outer surface area of the box, we first need to express the outer surface area as a function of x.

The outer surface area (A) of the box can be calculated by summing the areas of the base and the four sides.

The area of the base is given by A_base = x^2.

The area of each of the four sides is given by A_side = x * h.

Since the box has an open top, there is no top surface to consider.

Therefore, the total outer surface area can be expressed as A_total = A_base + 4 * A_side = x^2 + 4 * x * h.

We are given that the volume of the box is 108 cm^3, which can be expressed as V = x^2 * h.

Now, we can solve for h in terms of x:

h = V / x^2 = 108 / x^2.

Substituting this value of h into the expression for the total outer surface area:

A_total = x^2 + 4 * x * (108 / x^2) = x^2 + 432 / x.

To minimize the outer surface area, we need to find the value of x that minimizes A_total. To do this, we can take the derivative of A_total with respect to x and set it equal to zero:

dA_total / dx = 2x - 432 / x^2 = 0.

Simplifying this equation:

2x = 432 / x^2,

2x^3 = 432,

x^3 = 216,

x = cube root (216) = 6.

Therefore, the value of x that will produce the minimum outer surface area is 6 cm.

To find the minimum surface area, we substitute this value of x back into the expression for A_total:

A_total = (6)^2 + 432 / (6) = 36 + 72 = 108 cm^2.

Therefore, the minimum surface area of the box is 108 cm^2.

To find the value of x that will produce the minimum outer surface area, we can use a derivative.

Let's start with the formula for the volume of the box:
Volume = x^2 * h = 108 cm^3

Now, let's express the height in terms of x:
h = 108 / x^2

Next, let's find the formula for the outer surface area of the box:
Surface area = 2(x^2) + 4(xh)

Substitute h with its expression in terms of x:
Surface area = 2(x^2) + 4(x * (108 / x^2))
Surface area = 2x^2 + 4 * (108 / x)

Now, we can find the derivative of the surface area with respect to x:
d(Surface area)/dx = 4 - 432 / x^2

To find the minimum surface area, we need to find the critical points. Set the derivative equal to zero and solve for x:
4 - 432 / x^2 = 0
4x^2 = 432
x^2 = 432 / 4
x^2 = 108
x = sqrt(108)
x β‰ˆ 10.4 cm

So, the value of x that will produce the minimum outer surface area is approximately 10.4 cm.

To find the minimum surface area, substitute this value back into the surface area formula:
Surface area = 2(10.4^2) + 4(10.4 * (108 / (10.4)^2))
Surface area β‰ˆ 218.7 cm^2

Therefore, the minimum surface area is approximately 218.7 cm^2.