A box has a Square base. The Perimeter of the base plus the height is 120cm. What is the max volume of this box, and what are the dimensions of the this maximized box.

let the base be x by x cm

let the height be y
given: 4x + y = 120
y = 120 - 4x

V = x^2 y = x^2(120-4x)
= 120x^2 - 4x^3
dV/dx = 240x - 12x^2 = 0 for a max of V
12x^2 - 240x = 0
12x(x- 20) = 0
x = 0 ----> yields a minimum volume
or
x = 20

so the box is 20 cm by 20 cm by 40 cm for a max volume of 16000 cm^3

To find the maximum volume of the box, we need to determine the dimensions that would give us the largest possible volume.

Given that the box has a square base, let's assume the length of each side of the base is 'x'.

So, the perimeter of the base would be 4x, and the height of the box is h.

According to the given information, the perimeter of the base plus the height is 120cm. Therefore, we can write the equation:

4x + h = 120

To find the maximum volume, we need to express the volume of the box in terms of a single variable (either x or h). Since the base is square, the volume can be expressed as:

V = x^2 * h

To eliminate one variable, we can solve the equation 4x + h = 120 for h:

h = 120 - 4x

Now substitute this value of h in the volume equation:

V = x^2 * (120 - 4x)

Expand the equation:

V = 120x^2 - 4x^3

To find the maximum volume, we need to find the critical points of this equation. We can do this by finding the derivative of V with respect to x and setting it equal to zero:

dV/dx = 240x - 12x^2 = 0

Factor out x:

x(240 - 12x) = 0

Set each factor equal to zero:

x = 0 or 240 - 12x = 0

Solving the second equation:

240 - 12x = 0
12x = 240
x = 20

Since the length of a side cannot be zero, we discard the solution x = 0. Therefore, the only critical point is x = 20.

To determine if this point corresponds to a maximum volume, we can check the sign of the second derivative:

d^2V/dx^2 = -24x

Substitute x = 20:

d^2V/dx^2 = -24(20) = -480

Since the second derivative is negative, we can conclude that x = 20 corresponds to a maximum volume.

To find the dimensions of the maximized box, we know that the length of each side of the base, x, is 20cm. Substituting this value into the equation:

h = 120 - 4x
h = 120 - 4(20)
h = 120 - 80
h = 40

Therefore, the maximum volume of the box is given by:

V = x^2 * h
V = 20^2 * 40
V = 8000 cm^3

The dimensions of the maximized box are a square base with side length 20 cm and a height of 40 cm.

To find the maximum volume of the box with a square base, we need to optimize the dimensions of the box. We'll start by assigning variables to the unknowns in the problem.

Let's say the length of one side of the square base is x cm, and the height of the box is h cm.

Given that the perimeter of the base plus the height is 120 cm, we can set up the equation:

4x + h = 120

Next, we need to express the volume of the box in terms of x and h. Since the base is square, the area of the base is x^2. Therefore, the volume of the box is given by:

V = x^2 * h

To find the maximum volume, we can rewrite the equation for volume as:

V = x^2 * (120 - 4x)

Now, we can maximize the volume by finding the critical point where the derivative of V with respect to x is equal to zero.

dV/dx = 2x * (120 - 4x) + x^2 * (-4) = 0

Simplifying the equation, we get:

240x - 8x^2 - 4x^2 = 0

Rearranging the terms, we have:

12x^2 - 240x = 0

Factoring out 12x, we get:

12x(x - 20) = 0

So, we have two possible solutions: x = 0 or x = 20.

Since the dimensions of the box cannot be zero, we discard x = 0, leaving us with x = 20.

To find the height, we can substitute this value of x back into the equation we obtained from the perimeter:

4x + h = 120

4(20) + h = 120

80 + h = 120

h = 120 - 80

h = 40

Therefore, the dimensions of the box that maximize the volume are:
Length of one side of the square base = 20 cm
Height = 40 cm

Now, we can calculate the maximum volume of the box by substituting these dimensions into the volume equation:

V = x^2 * h

V = 20^2 * 40

V = 400 * 40

V = 16,000 cm^3

So, the maximum volume of the box is 16,000 cm^3, and the dimensions of the maximized box are:
Length of one side of the square base = 20 cm
Height = 40 cm