Question

Find the dimensions of a box with a square base with surface area 44 and the maximal volume. (Use symbolic notation and fractions where needed.)

side of base
height
maximum volume

Answer 1

To find the dimensions of the box with a square base and maximum volume, we can use the given surface area of 44 and solve for the side length of the base, height, and maximum volume.

Step-by-step explanation:

To find the dimensions of a box with a square base that has a surface area of 44 units, we first need to find the side length of the square base.

Let's call it 's'. Since the surface area of a square is given by the formula A = 6s^2, we can rearrange the formula to solve for 's'. 6s^2 = 44. Divide both sides by 6 to get s^2 = 7. Simplifying further, s = √7, which is the side length of the square base.

Next, to find the height of the box, we need to determine the scale factor. The problem states that the dimensions of the larger square are twice that of the first square. This means the height of the box is also doubled. So, the height is 2 * √7.

To find the maximum volume, we use the formula V = s^2 * h. Substituting the values we found, the maximum volume of the box is (√7)^2 * 2 * √7, which simplifies to 14 units.

Answer 2

The maximal volume of the box with a square base and surface area 44 is
`\((52)/(6)\)` cubic units.

Let's denote the side length of the square base as
`\(s\)` and the height of the box as
`\(h\)`. Given that the surface area of the box is 44, we can set up an equation using the formula for the surface area of a box:

`Surface Area = \(2 * \text{base area} + \text{perimeter of base} * \text{height}\)`

For a square base, the base area is
`\(s^2\)` and the perimeter is
`\(4s\)`. The surface area given is 44, so:

`\[2s^2 + 4sh = 44\]`

Now, we want to express the volume of the box in terms of
`\(s\)` and
`\(h\)`. The volume of a box is given by the formula:

`Volume = \(\text{base area} * \text{height}\)`

For a square base, the base area is
`\(s^2\)`, so the volume is
`\(s^2 * h\)`.

The goal is to find the dimensions that maximize the volume. We can use the fact that the side length of the base is given as
`\(\left((52)/(6)\right)^{(1)/(2)}\)` to determine the dimensions that result in the maximal volume.

Let's solve for
`\(s\)` and
`\(h\)` using the surface area equation and then find the maximum volume:

Given:
`\(s = \left((52)/(6)\right)^{(1)/(2)}\)`

`\[2s^2 + 4sh = 44\]`

`\[2 * \left((52)/(6)\right) + 4 * \left((52)/(6)\right)^{(1)/(2)} * h = 44\]`

`\[(104)/(6) + 4 * \left((52)/(6)\right)^{(1)/(2)} * h = 44\]`

`\[4 * \left((52)/(6)\right)^{(1)/(2)} * h = 44 - (104)/(6)\]`

`\[4 * \left((52)/(6)\right)^{(1)/(2)} * h = (176)/(6)\]`

`\[h = \frac{(176)/(6)}{4 * \left((52)/(6)\right)^{(1)/(2)}}\]`

`\[h = \frac{44}{\left((52)/(6)\right)^{(1)/(2)}}\]`

So, the height of the box is
`\(\left((52)/(6)\right)^{(1)/(2)}\)`.

Now, let's calculate the volume:

`\[V = s^2 * h\]`

`\[V = \left((52)/(6)\right)^{(1)/(2)} * \left((52)/(6)\right)^{(1)/(2)}\]`

`\[V = (52)/(6)\]`

Therefore, The answer is
`\((52)/(6)\)` cubic units.

The complete question is here:

Find the dimensions of a box with a square base with surface area 44 and the maximal volume.

(Use symbolic notation and fractions where needed.)