Given a 30cm by 30cm sheet of metal has squares of side x cm cut from each corner. Calc. the dimension for the greatest volume.

well, the cutouts on each corner have dimensions x times x, or area x^2

So the box has dimensions. So the dimensions of the box will be LxWxH (length, width, height).

30=L+2x (x is the height)
30=W+2xfrom the metal dimensions.

so L=30-2x
W=30-2x

Volume=LWH= x(30-2x)^2
take the derivative, and set to zero.

dV/dx=0= you do it. Solve for x

To find the dimensions for the greatest volume, we can apply calculus and optimize the volume function.

Let's denote the side length of the squares as x cm. This means that the length and width of the resulting metal tray will be (30 - 2x) cm, as the x cm squares are cut from each corner.

The height of the tray will be x cm.

The volume V of the tray can be calculated using the formula:

V = length * width * height

Substituting the values, we have:

V = (30 - 2x) * (30 - 2x) * x

Expanding the equation, we get:

V = (900 - 60x - 60x + 4x^2) * x
V = 4x^3 - 120x^2 + 900x

To find the dimensions for the greatest volume, we need to find the value of x that maximizes the volume V.

To do so, we can take the derivative of V with respect to x and set it equal to zero. Then, solve for x.

dV/dx = 0

Taking the derivative of V with respect to x:

dV/dx = 12x^2 - 240x + 900

Setting it equal to zero:

12x^2 - 240x + 900 = 0

Now, we solve this quadratic equation for x.

By factoring or applying the quadratic formula, we find that x = 5 cm or x = 15 cm.

To determine which value of x gives the maximum volume, we can evaluate the second derivative of V with respect to x.

d^2V/dx^2 = 24x - 240

Substituting x = 5 cm and x = 15 cm, we get:

d^2V/dx^2 (x = 5 cm) = 24(5) - 240 = -120
d^2V/dx^2 (x = 15 cm) = 24(15) - 240 = 0

Since the second derivative is negative when x = 5 cm, this corresponds to a maximum volume.

Therefore, the dimensions that result in the greatest volume are:
- The length and width of the metal sheet after cutting the squares will be 30 - 2(5) = 20 cm.
- The height of the resulting tray will be 5 cm.

Hence, the dimensions for the greatest volume are a 20cm by 20cm by 5cm metal tray.

To find the dimensions that will result in the greatest volume, we need to perform the following steps:

Step 1: Define the variables:
Let x represent the side length of the squares cut from each corner.
Let L represent the length of the metal sheet after the squares are cut.
Let W represent the width of the metal sheet after the squares are cut.
Let V represent the volume of the resulting box.

Step 2: Determine the dimensions of the metal sheet after cutting:
Since squares with side length x are cut from each corner of the original 30cm by 30cm metal sheet, the new dimensions will be:
Length = original length - 2 * x
Width = original width - 2 * x

L = 30cm - 2x
W = 30cm - 2x

Step 3: Calculate the volume of the resulting box:
The volume of a rectangular box can be calculated by multiplying its length, width, and height.
In this case, the height is the side length of the squares cut from each corner, which is also x.

V = L * W * x

Step 4: Express the volume in terms of a single variable:
Substitute the expressions for L and W from step 2 into the equation for V:

V = (30cm - 2x) * (30cm - 2x) * x

Step 5: Simplify the equation:
Expand and simplify the equation:

V = (900cm² - 60cmx - 60cmx + 4x²) * x
V = (900cm² - 120cmx + 4x²) * x
V = 900cm²x - 120cmx² + 4x³

Step 6: Find the critical points:
To find the critical points, we need to find the values of x where the derivative of V with respect to x equals zero.

Differentiate the equation for V with respect to x:

V' = 900cm² - 120cmx + 4x³

Set V' equal to zero and solve for x:

0 = 900cm² - 120cmx + 4x³

Step 7: Solve for x:
We need to solve the equation 0 = 900cm² - 120cmx + 4x³ for x. This can be done using numerical methods or a graphing calculator.

The value(s) of x obtained from solving the equation will give us the dimensions that result in the greatest volume.