Postal regulations specify that a parcel sent by priority mail may have a combined length and girth of no more than 144 in. Find the dimensions of a rectangular package that has a square cross section and largest volume that may be sent by priority mail. Hint: The length plus the girth is 4x + l.

length
width
height

What is the volume of such a package?

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To find the dimensions of a rectangular package with a square cross-section that has the largest volume allowed by the postal regulations, we can use optimization techniques.

Let's call the side length of the square cross-section "x". Since the package has a square cross-section, the width and height of the package will also be equal to "x".

According to the postal regulations, the combined length and girth of the package should be no more than 144 inches. The girth of a rectangular package is equal to twice the sum of the width and height, which in this case is 2x + 2x = 4x. So, the length plus the girth is 4x + l.

According to the regulations, the length plus the girth should be equal to or less than 144 inches:

4x + l ≤ 144

Now, we need to express the volume of the package in terms of "x" and maximize it. The volume of a rectangular package is given by V = length × width × height.

In this case, since the width and height are equal to "x", the volume can be expressed as:

V = l × x × x = l × x^2

We want to find the largest possible volume, so we need to maximize V. To do this, we need to express V as a function of a single variable.

We can use the inequality constraint 4x + l ≤ 144 to express "l" in terms of "x":

l ≤ 144 - 4x

Substituting this expression for "l" into the volume equation:

V = (144 - 4x) × x^2

Now, we have expressed the volume of the package as a function of a single variable "x". To find the largest possible volume, we need to find the maximum value of this function.

To do this, we can take the derivative of the volume function with respect to "x":

dV/dx = 144x - 8x^2

Setting the derivative equal to zero to find the critical points:

144x - 8x^2 = 0

Simplifying the equation:

8x(18 - x) = 0

This equation has two solutions: x = 0 and x = 18. Since the side length cannot be zero, we can ignore the x = 0 solution.

Now, we have found the critical point x = 18. To confirm that this is the maximum, we can determine the second derivative of the volume function:

d^2V/dx^2 = 144 - 16x

Evaluating the second derivative at the critical point x = 18:

d^2V/dx^2 = 144 - 16(18) = -144

Since the second derivative is negative, we can conclude that x = 18 corresponds to a maximum.

Therefore, the side length of the square cross-section of the package is 18 inches. The length, width, and height would also be equal to 18 inches.

To find the volume of the package, we can substitute the side length into the volume equation:

V = (144 - 4(18)) × 18^2 = 144 × 18^2 = 46,656 cubic inches.

Therefore, the volume of the package is 46,656 cubic inches.