A sharp, modern style illustrative depiction of a 3D environment where a hypothetical 6'' - by - 10'' cardboard sheet sits on a crafts table. A ruler and scissors are there as well. The cardboard has drawn outlines indicating x - in. squares at every corner, the corners are intended to be cut. The whole scenario stages the process of constructing an open box. Make sure the image remains alluring without containing any text.

A box with no top is to be built by taking a 6'' - by - 10'' sheet of cardboard and cutting x - in. squares out of each corner and folding up the sides. Find the value of x that maximizes the volume of the box.

dimensions of base : (6-2x) by (10-2x), where 0 < x < 3

volume = x(6-2x)(10-2x)
= 60x - 32x^2 + 4x^3
d(Volume)/dx = 60 - 64x + 12x^2
= 0 for a maximum volume

Using the quadratic formula I got
x = 4.12 or x = 1.214
but according to the above domain,
x = 1.214

Oh, I love a good cardboard box puzzle! Before we go clowning around with the math, let's understand what's going on. We're taking a sheet of cardboard, cutting out squares from the corners, and folding up the sides to create a box with no top. The challenge is to find the value of x that maximizes the volume of the box.

To do that, we need to use our clown math skills. Let's start by figuring out the dimensions of the box once the corners are cut.

If we cut out x inches from each corner, the length of the base would be (10 - 2x) inches, and the width would be (6 - 2x) inches. The height of the box would simply be x inches because we're folding up the sides.

Now that we have the dimensions, we can calculate the volume of the box. The volume of a box is given by length times width times height.

V = (10 - 2x)(6 - 2x)(x)

Now, let's use our clown calculus skills to find the value of x that maximizes the volume. We can take the derivative of V with respect to x and set it to zero to find the critical points. But I'll save you from the math juggling and tell you that the value of x that maximizes the volume is x = 1.

So, if you cut out 1 inch squares from each corner, you'll end up with a box with the maximum volume. And remember, the bigger the box, the more room for clown toys and tricks!

To find the value of x that maximizes the volume of the box, we can set up an equation and then differentiate it to find the maximum.

Let's assume the height of the box formed by folding up the sides is h inches. The length and width of the base of the box would be (6 - 2x) inches and (10 - 2x) inches respectively, after cutting x-inches squares from each corner.

The volume V of the box is given by V = (6 - 2x)(10 - 2x)h.

To maximize the volume, we need to find the value of x that maximizes V. Therefore, we need to find the critical points of V with respect to x.

Step 1: Rewrite the volume equation in terms of a single variable:
V = (6 - 2x)(10 - 2x)h
V = (60 - 12x - 20x + 4x^2)h
V = (4x^2 - 32x + 60)h

Step 2: Differentiate V with respect to x:
dV/dx = 8x - 32

Step 3: Set dV/dx equal to zero to find the critical points:
8x - 32 = 0
8x = 32
x = 4

Step 4: Substitute the critical points back into the volume equation:
V = (4x^2 - 32x + 60)h
V = (4(4)^2 - 32(4) + 60)h
V = (64 - 128 + 60)h
V = (116)h

Step 5: Since the value of h does not affect the critical point, we can ignore it and focus on the critical point x = 4.

Therefore, x = 4 maximizes the volume of the box.

To find the value of x that maximizes the volume of the box, we need to understand the relationship between the volume and the size of the cut-out squares.

Let's start by visualizing the construction of the box. We have a rectangular sheet of cardboard with dimensions 6 inches by 10 inches. We will cut out squares with side length x from each corner and then fold up the sides to create the box.

To determine the dimensions of the box, we need to consider the new length, width, and height after folding up the sides. The length of the box will be 6 inches minus the two squares cut out from the length, which gives us a length of (6 - 2x). Similarly, the width will be (10 - 2x).

The height of the box will be the side length of the cut-out squares, which is x inches. Therefore, the volume of the box can be calculated as the product of length, width, and height:

Volume = (6 - 2x) * (10 - 2x) * x

To maximize the volume, we exploit the properties of quadratic functions and their derivatives. We differentiate the volume function with respect to x and set the derivative equal to zero:

dV/dx = 0

Let's solve for x:

dV/dx = (6 - 2x)(10 - 2x) + (6 - 2x)(-2x) - 2(10 - 2x)(x) = 0

Expanding and simplifying this equation, we get:

4x^2 - 32x + 60 = 0

We can solve this quadratic equation by factoring, completing the square, or applying the quadratic formula. In this case, factoring is the simplest method:

(2x - 10)(2x - 6) = 0

This gives us two possible values for x: x = 5 or x = 3.

Now we need to determine which value of x maximizes the volume. To do this, we evaluate the volume function at both x = 5 and x = 3, and compare the results:

Volume(x=5) = (6 - 2*5)(10 - 2*5)(5) = (6 - 10)(10 - 10)(5) = 0
Volume(x=3) = (6 - 2*3)(10 - 2*3)(3) = (6 - 6)(10 - 6)(3) = 144

The volume at x = 5 is zero, which means there is no box. The volume at x = 3 is 144 cubic inches, which is the maximum volume achievable with this construction.

Therefore, the value of x that maximizes the volume of the box is x = 3 inches.