Postal regulations specify that a parcel sent by priority mail may have a combined length and girth of no more than 96 in. Find the dimensions of a rectangular package that has a square cross section and the largest volume that may be sent via priority mail. Hint: The length plus the girth is 4x + h (see the figure below).

well, we know

4x+h = 96, so h=96-4x
v = hx^2
= x^2(96-4x)
= 96x^2 - 4x^3
= 4(24x^2 - x^3)

dv/dx = 4(48x - 3x^2)
= 12x(16 - x)

dv/dx = 0 when x = 0 or 16
so, h = 96-4*16 = 32

So, if the package is 16x16x32
v = 8192 in^3

Now, technically, 4x+h <= 96, but it should be clear that max volume occurs when 4x+h is as large as possible.

Well, if we want to find the largest volume that can be sent via priority mail, we'll need to maximize the dimensions of the package within the given constraints.

Let's call the side length of the square cross section "x" and the height of the package "h". Since the length plus the girth is given as 96 inches, we can write the equation as:

4x + h = 96

Now, to find the largest volume, we need to maximize the product of the dimensions. The volume of a rectangular package is given by V = x^2 * h.

To get rid of the "h" variable, let's rewrite the equation as h = 96 - 4x.

Substituting this into the volume equation, we get V = x^2 * (96 - 4x).

Now, let the humor commence! Let's apply some clown logic here...

We know that the dimension x has to be positive since we're dealing with lengths. So, to maximize the volume, we'll need to find the maximum point of this quadratic equation.

To find the maximum point, we can use some fancy circus math and take the derivative of the volume equation:

dV/dx = 2x(96 - 4x) - x^2 * 4

Simplifying, we get:

dV/dx = 192x - 8x^2 - 4x^2

Combining like terms, we have:

dV/dx = -12x^2 + 192x

Set this equal to zero and solve for x to find the critical points:

-12x^2 + 192x = 0

12x(-x + 16) = 0

So, we have two solutions: x = 0 and x = 16.

Now, back to reality! Since x represents a length, x = 0 doesn't make sense. We'll ignore that solution and focus on x = 16.

Therefore, the side length of the square cross section is 16 inches. But we still need to find the height of the package (h).

Using the equation 4x + h = 96 and plugging in x = 16, we can solve for h:

4(16) + h = 96

64 + h = 96

h = 96 - 64

h = 32

So, the dimensions of the rectangular package that has the largest volume that can be sent via priority mail are:

Side length of square cross section: 16 inches
Height of the package: 32 inches

To find the dimensions of a rectangular package with a square cross section and the largest volume that can be sent via priority mail, we need to consider the restrictions given by the postal regulations.

Let's assume the side length of the square cross section is "x". Therefore, the length of the rectangular package (which is equal to the side length of the square cross section) is also "x", and the width and height are both "x".

The formula for the combined length and girth is given as 4x + h, where "h" represents the height of the package. In this case, the combined length and girth should be less than or equal to 96 inches.

So, we have the equation:

4x + h ≤ 96

Substituting "x" for "h" (since the height is also "x"), we get:

4x + x ≤ 96

Combining like terms:

5x ≤ 96

Dividing both sides of the inequality by 5:

x ≤ 19.2

Since "x" represents the side length of the square cross section, we can conclude that the side length should be less than or equal to 19.2 inches.

To maximize the volume, we need to choose the largest possible side length. However, since the side length must be a whole number (since it represents inches), we round down the value of 19.2 to the nearest whole number, which is 19.

Therefore, the dimensions of the rectangular package with a square cross section that has the largest volume and can be sent via priority mail are:

Length = Width = Height = 19 inches.

To find the dimensions of a rectangular package that has a square cross section and the largest volume that may be sent via priority mail, we need to maximize the volume while considering the constraint that the combined length and girth should not exceed 96 inches.

Let's break down the problem and solve step by step:

Step 1: Define the variables
Let's assume that the side length of the square cross-section is "x" inches, and the height of the rectangular package is "h" inches.

Step 2: Define the volume
The volume of a rectangular package with a square cross-section is given by:
Volume = length * width * height

In this case, the length and width are both "x" inches (since it's a square cross-section), and the height is "h" inches. So the volume is:
Volume = x * x * h = x^2 * h

Step 3: Define the constraint
The combined length and girth should not exceed 96 inches. The length is x inches and the girth is 2(x + h) inches. So the constraint is:
Length + Girth ≤ 96

Substituting the values, we get:
x + 2(x + h) ≤ 96
x + 2x + 2h ≤ 96
3x + 2h ≤ 96

Step 4: Determine the maximum volume
To maximize the volume, we need to find the maximum value of x^2 * h, while satisfying the constraint 3x + 2h ≤ 96.

One approach is to solve the constraint equation for h, in terms of x:
3x + 2h = 96
2h = 96 - 3x
h = (96 - 3x) / 2

Substituting this value of h in the volume equation, we get:
Volume = x^2 * ((96 - 3x) / 2)

Step 5: Find the maximum value of the volume
To find the maximum value, we can take the derivative of the volume equation with respect to x and set it equal to 0.

d/dx (Volume) = d/dx (x^2 * ((96 - 3x) / 2))
= (2x * ((96 - 3x) / 2)) + (x^2 * (-3/2))
= (x * (96 - 3x)) - (3x^2 / 2)
= 96x - 3x^2 - (3x^2 / 2)

Setting it to 0 and solving for x:
96x - 3x^2 - (3x^2 / 2) = 0
Multiplying by 2 to get rid of the denominator:
192x - 6x^2 - 3x^2 = 0
Combining like terms:
192x - 9x^2 = 0
Factor out common factor:
x(192 - 9x) = 0
Setting each factor to 0 and solving for x:
x = 0 (non-negative value)
192 - 9x = 0
9x = 192
x = 192/9
x = 21.33 inches (approx.)

Since the side length of the square cannot be in decimal values, we round down the value to 21 inches.

Now that we have the side length, we can substitute it back into the constraint equation to find the value of h:
3x + 2h = 96
3(21) + 2h = 96
63 + 2h = 96
2h = 96 - 63
2h = 33
h = 33/2
h = 16.5 inches (approx.)

So the dimensions of the rectangular package with the largest volume that can be sent via priority mail are approximately: 21 inches (side length of the square cross-section) and 16.5 inches (height of the rectangular package).

Postal regulations specify that a parcel sent by priority mail may have a combined length and girth of no more than 120 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
in
width
in
height
in

What is the volume of such a package?

in3