Suppose an airline policy states that all the baggage must be boxed shaped with a sum of length, width, and height not exceeding 138 inches. What are the dimensions and volume of a square based box with the greatest volume under these conditions.

let the dimensions by x by x by y

we know x+x+y ≤ 138
2x + y ≤ 138
y ≤ 138-2x

V = x^2 y = x^2(138-2x)
= 138x^2 - 2x^3
dV/dx = 276x - 6x^2
= 0 for a max of V
6x^2 = 276x
6x = 276
x = 46
then y = 138-92 = 46

so the box must be a cube of 46 by 46 by 46
as suspected

notice 46+46+46 ≤ 138

Well, I must say, that's a rather square question! So, let's solve this puzzle together, shall we?

If we have a square base, it means that the length and width are the same, let's call it "x." Now, we need to find the height, but we know that the sum of length, width, and height must not exceed 138 inches. So, we have the equation:

2x + h ≤ 138

Now, to maximize the volume, we need to consider the equation for the volume of a box: V = L x W x H. In this case, V = x^2 x h.

Substituting 2x + h for 138, we get:

2x + h = 138
h = 138 - 2x

Now, let's substitute this value of h in the volume equation:

V = x^2 x (138 - 2x)

To find the maximum volume, we can take the derivative of V with respect to x and set it to zero:

dV/dx = 138x - 4x^2 = 0

Simplifying this equation gives us:

4x^2 - 138x = 0
2x(2x - 69) = 0

So, we have two possible values for x: x = 0 and x = 69/2. Since we can't have a zero-length box, we discard x = 0.

Therefore, x = 69/2.

Substituting this value back into h = 138 - 2x, we find:

h = 138 - 2(69/2) = 0

Uh-oh! Looks like something went wrong. It seems we've reached the end of our clownish calculations, and it appears that there's no feasible solution given these conditions. It seems we can't create a box with the greatest volume using a square base and a total sum of 138 inches.

But don't worry! The airline policies surely allow for other shapes and dimensions. So, you'll still be able to pack your bags and fly happily! Safe travels!

To find the dimensions and volume of a square-based box with the greatest volume under the given conditions, we can use optimization techniques.

Let's assume the length, width, and height of the box are x, x, and y, respectively.

According to the airline policy, the sum of the length, width, and height should not exceed 138 inches. Therefore, we have the equation:

x + x + y = 138 (Equation 1)

The volume of a square-based box is given by the formula:

Volume = length * width * height = x * x * y = x^2 * y

We need to maximize the volume, so we can express it as a function of a single variable, either x or y. Substituting the value of x from Equation 1, we can rewrite the volume equation as:

Volume = (138 - y)^2 * y (Equation 2)

To maximize the volume, we can differentiate Equation 2 with respect to y, set it equal to zero, and solve for y.

dV/dy = 2(138 - y)(-1)y + (138 - y)^2 = 0

Simplifying the above equation, we get:

-2y(138 - y) + (138 - y)^2 = 0

Expanding and rearranging, we have:

-2y^2 + 276y + y^2 - 276y + 138^2 - 138y = 0

Simplifying further, we get:

-y^2 + 138^2 = 0

Solving for y, we find:

y = 138

Now, substituting the value of y back into Equation 1, we can solve for x:

x + x + 138 = 138 (since y = 138)

2x = 0

x = 0

Therefore, if y = 138, the value of x is 0, which is not physically possible.

Hence, it appears that there is no square-based box that meets the given airline policy conditions with the greatest volume.

To find the dimensions and volume of a square-based box with the greatest volume under the given conditions, we can approach it using mathematical optimization.

1. Let's start by defining the variables:
- Let "s" represent the length of one side of the square base.
- Let "h" represent the height of the box.

2. We need to consider the constraints provided by the airline policy:
- The sum of the length, width, and height of the box should not exceed 138 inches.
- The length of the box's diagonal should also not exceed 138 inches.

3. Since we have a square base, the length, width, and diagonal can be expressed in terms of "s":
- The length (l) and width (w) of the box are both equal to "s".
- The diagonal (d) of the base can be found using the Pythagorean theorem: d = √(s^2 + s^2) = √2s.

4. Based on the constraints, we can write the following equations:
- s + s + h ≤ 138
- √2s ≤ 138

5. We can simplify the first equation: 2s + h ≤ 138. Solving for h: h ≤ 138 - 2s.

6. Substitute h in the second equation: √2s ≤ 138 - 2s.

7. Square both sides of the inequality: 2s ≤ (138 - 2s)^2.

8. Expand the right side: 2s ≤ 138^2 - 2 * 138 * 2s + 4s^2.

9. Simplify the equation: 0 ≤ 138^2 - 4s^2 - 2 * 138 * 2s.

10. Rearrange the equation, moving all terms to one side: 4s^2 + 4 * 138 * 2s - 138^2 ≥ 0.

11. Calculate the discriminant of the quadratic equation above: Δ = (4 * 138 * 2)^2 - 4 * 4 * (-138^2).

12. Simplify the discriminant: Δ = 4^2 * 138^2 * (2^2 + 4) = 4^2 * 138^2 * 20.

13. Since the discriminant is positive, there will be real solutions for "s".

14. Find the positive solution for "s" by using the quadratic formula: s = (-4 * 138 * 2 + √Δ) / (2 * 4).

15. Calculate the value of "s" based on the formula.

16. With the value of "s", you can find the height using the equation h ≤ 138 - 2s.

17. Calculate the height based on the formula.

18. Finally, you can calculate the volume of the box using the formula: volume = s^2 * h.

These steps will give you the dimensions and volume of a square-based box with the greatest volume under the given conditions.