The postal service places a limit of 84 inches on the combined lenght and girth (distance around) of a package to be sen parcel post. What Dimensions of a rectangular box with square cross-section will contain the largest volume that can be mailed?
4x+y = 84, so
y = 84-4x
v = x^2 y = x^2(84-4x) = 84x^2 - 4x^3
dv/dx = 168x - 12x^2
= 12x(14-x)
dv/dx=0 when x=14
so the box is 14 x 14 x 28
Oh, you're trying to think outside the box? Well, how about we try to think inside the box instead?
Now, let's start with a square cross-section, with each side measuring 'x'. Since it's a rectangular box with a square cross-section, the length, width, and height of the box will all be 'x'.
To calculate the volume, we multiply the length, width, and height together. So the volume of the box is: V = x * x * x = x^3.
Now, we need to consider the constraint given by the postal service. The combined length and girth must be less than or equal to 84 inches.
The girth of a rectangular box is 2 * (length + width). In our case, the girth would be 2 * (x + x) = 4x.
Adding the length to the girth, we get a total distance around the box: x + 4x = 5x.
According to the postal service, this total distance around must be less than or equal to 84 inches. So, we can write the inequality: 5x ≤ 84.
Now, we can solve this inequality for 'x' to find the maximum allowable value. After solving, we get x ≤ 16.8.
Since we cannot have a fraction of an inch as a side length, we round down to x = 16 inches.
Therefore, the dimensions of the rectangular box with a square cross-section that will contain the largest volume that can be mailed is 16 inches on each side.
To find the dimensions of a rectangular box with a square cross-section that would contain the largest volume that can be mailed, we need to maximize the volume while considering the given limitation of 84 inches on the combined length and girth.
Let's assume the length, width, and height of the box are all equal to x.
The length of the box (x) will be one side, and the combined length and girth will be the sum of the other three sides: 2x + 2x = 4x.
According to the given limitation, the combined length and girth should not exceed 84 inches:
4x ≤ 84
Simplifying the inequality:
x ≤ 84/4
x ≤ 21
Since the dimensions of a rectangular box with a square cross-section are the same, x should be less than or equal to 21 inches.
Now, let's calculate the volume of the box:
Volume = length × width × height
Volume = x × x × x
Volume = x^3
To maximize the volume, we need to find the value of x that yields the largest result for x^3.
Since the problem is now reduced to maximizing a single variable, we can differentiate the equation and solve for x.
dV/dx = 3x^2
Setting the derivative equal to zero to find the critical points:
3x^2 = 0
x^2 = 0
x = 0
Since the length cannot be zero, we can disregard this solution.
Now, we need to check the endpoints of the interval [0, 21] to see if they yield any maximum values:
When x = 0, the volume is 0.
When x = 21, the volume is 21^3 = 9,261 cubic inches.
Therefore, the maximum volume that can be mailed with a rectangular box with a square cross-section is 9,261 cubic inches, and the dimensions of the box are 21 inches in length, width, and height.
To find the dimensions of a rectangular box with a square cross-section that will contain the largest volume under the given constraints, we can use calculus and the concept of optimization.
Let's assume the dimensions of the rectangular box as follows:
Length = L
Width = W
Height = H
Since the box has a square cross-section, the width and height are equal (W = H). The length encloses the other three sides and forms the girth (circumference) of the package. Therefore, the girth can be represented as follows:
Girth = 2(W + H) = 2(2W) = 4W
From the given constraint, the limit on the combined length and girth is 84 inches:
Length + Girth = L + 4W ≤ 84
Next, we need to express the volume of the box in terms of a single variable. Since the box has a square cross-section, the volume is given by:
Volume = Length * Width * Height = L * W * W = L * W^2
To proceed with optimization, we need to represent the volume in terms of a single variable. We can use the constraint equation to express one variable in terms of the other.
L = 84 - 4W
Substituting this value of L in the volume equation:
Volume = (84 - 4W) * W^2
Now, we have the volume equation as a function of a single variable, W. To find the maximum volume, we can take the derivative of the function with respect to W and set it equal to zero.
d(Volume)/dW = 0
Simplifying this derivative equation, we get:
12W^2 - 336W + 1680 = 0
Solving this quadratic equation, we find two values of W: W = 4 and W = 35.
However, we need to check if these critical values are within the constraints. Using the constraint:
Length + Girth = L + 4W ≤ 84
For W = 4:
Length + Girth = 84 - 16 = 68 (valid)
For W = 35:
Length + Girth = 84 - 140 = -56 (invalid)
Therefore, the valid value for W is 4. Now, we can find the corresponding value of L using the constraint equation:
Length = 84 - 4W = 84 - 16 = 68
Thus, the dimensions of the rectangular box with the largest volume that can be mailed are:
Length = 68 inches
Width/Height = 4 inches
We can then calculate the volume of the box using these dimensions:
Volume = Length * Width^2 = 68 * 4^2 = 68 * 16 = 1088 cubic inches
Hence, a rectangular box with dimensions 68 inches (length) x 4 inches (width and height) will contain the largest volume that can be mailed under the given constraints.