# A square sheet of cardboard 18 inches is made into an open box (there is no top), by cutting squares of equal size out of each corner and folding up the sides. Find the dimensions of the box with the maximun volume.

Volume= base(width)height

but base + 2H =18

base + 2W=18

since squares are cut out of the corners, base and width are the same.

Volume=base^2 * height

but height= 9 - base/2

Volume = base^2 (9-base/2)

differentiate, set to zero, solve.

oooops you are in precal. If you don't know how to differentiat a uv function, then

graph this function for max volume.

x^4+6<5X^2

## -18X

The maximum volume of the box is when the base is 6 inches and the height is 3 inches.

## To find the dimensions of the box with the maximum volume, we can start by drawing a diagram or visualizing the situation.

We have a square sheet of cardboard, and we need to cut squares of equal size out of each corner. Let's represent the side length of these squares as "x".

When we fold up the sides, the resulting shape will be a box with an open top. The base of the box will have dimensions: (18 - 2x) x (18 - 2x), since we are cutting squares of size "x" from each corner.

The height of the box can be determined by the remaining length of the cardboard after cutting the squares. Since there are two sets of squares being cut (one from each side), the height will be 9 - (x/2).

Now we can calculate the volume of the box. It is given by the formula: Volume = base^2 * height.

Substituting the dimensions we calculated, the volume becomes: V = (18 - 2x)^2 * (9 - x/2).

To find the dimensions of the box with the maximum volume, we can use calculus by differentiating the volume function and setting it to zero. However, since you are in precalculus, we can use a graphical approach.

You also mentioned another inequality: x^4 + 6 < 5x^2. This inequality seems unrelated to the original question about the dimensions of the box. Please provide more context or clarify if you have any specific questions about it.

## To find the dimensions of the box with maximum volume, we need to find the value of the base that will maximize the volume. Let's start by writing down the equations:

The equation for the volume of the box is given by: V = base^2 * height

We know that base + 2h = 18, where h is the height of the box.

Since squares are cut out of the corners, the base and width are the same: base = width.

Substituting h = (9 - base/2) into the volume equation, we get:

V = base^2 * (9 - base/2) = 9base^2 - (base^3)/2

To find the maximum volume, we need to find the value of base that maximizes this function.

Since you are in precalculus and may not be familiar with differentiation, we can find the maximum volume by graphing the function.

The equation you provided, x^4 + 6 < 5x^2, seems to be unrelated to the problem of finding the maximum volume of the box. Could you please provide the correct equation that relates to the box volume optimization?