1. A rectangular package to be sent by a postal service can have a maximum combined length and girth (perimeter of a cross section) of 108 inches. Find the dimensions of the package of maximum volume that can be sent. (Assume the cross section is square.)
If the package has a combined length and girth of 108 inches. We can determine that:
4x + y = 108
We are also know that the cross section is square, so:
let V=volume
V = x^2y
Let's solve for y in the first equation and plug it into the second one:
y = 108 - 4x
We have
V = x^2(108 - 4x)
V = 108x^2 - 4x^3
In order to solve for the maximum is when V' = 0
V' = 216x - 12x^2 = 0
x(216 - 12x) = 0
Therefore, we have 2 solutions,
first one is x = 0
And this is the one we will work on:
216 - 12x = 0
12x = 216
x = 18
x = 0 is when the package is a minimum, so the maximum occurs when x = 18. Now let's solve for y
Just plug in the value of x in the first equation:
y = 108- 4x = 108- 4(18) = 36
The answers are:
x = 18 inches
y = 36 inches
To find the dimensions of the package of maximum volume, we can use calculus. Let's denote the length and width of the package as L and W, respectively.
Given that the combined length and girth should be a maximum of 108 inches, we can write the equation:
2L + 2W ≤ 108
Also, assuming the cross-section is square, we have L = W.
Substituting L = W into the equation, we get:
4L ≤ 108
Dividing both sides by 4, we have:
L ≤ 27
Since the dimensions of a package cannot be negative, the range of possible values for L is 0 ≤ L ≤ 27.
Now, let's express the volume of the package in terms of L:
V = L * L * W
Substituting L = W, we get:
V = L^2 * L
V = L^3
To find the maximum volume, we need to find the critical points of the volume function. Taking the derivative of V with respect to L, we have:
dV/dL = 3L^2
Setting dV/dL = 0 and solving for L, we get:
3L^2 = 0
L = 0
However, L = 0 is not within the range of possible values for L, so it is not a valid solution.
We can also check the endpoints of the range, L = 0 and L = 27:
V(0) = 0^3 = 0
V(27) = 27^3 = 19683
Therefore, the maximum volume that can be sent is when L = 27.
Thus, the dimensions of the package with maximum volume are:
Length (L) = 27 inches
Width (W) = 27 inches
To find the dimensions of the package of maximum volume that can be sent, we need to determine the dimensions that will maximize the volume of the package while still satisfying the given constraint on the combined length and girth.
Let's assume the length of the rectangular package is "L" and the width and height are both "w".
The combined length and girth of the package can be expressed as:
Length + 2*(width + height)
Since the cross-section is assumed to be square, we have:
width = height = w
Therefore, the combined length and girth becomes:
L + 2*(w + w) = L + 4w
According to the given constraint, the combined length and girth of the package should be 108 inches:
L + 4w = 108
Now, let's express the volume of the package, V, in terms of L and w:
V = L * w * w = Lw^2
To proceed further, we need to eliminate L from the equation by rearranging the constraint equation and solving for L:
L = 108 - 4w
Substituting this value of L in the volume equation:
V = (108 - 4w) * w^2
Now, we have the volume of the package in terms of a single variable, w. To find the dimensions of the package of maximum volume, we need to find the value of w that maximizes the volume.
To do that, we can take the derivative of the volume function with respect to w, set it equal to zero, and solve for w.
dV/dw = 0
Differentiating V = (108 - 4w) * w^2 with respect to w, we get:
dV/dw = -8w^2 + 216w
Setting this equal to zero and solving for w:
-8w^2 + 216w = 0
w(-8w + 216) = 0
Since the width of the package cannot be zero, we solve the equation -8w + 216 = 0:
-8w = -216
w = -216 / -8
w = 27
Now that we have the value of w, we can substitute it back into the constraint equation to find the corresponding value of L:
L + 4w = 108
L + 4(27) = 108
L + 108 = 108
L = 0
Therefore, the dimensions of the package of maximum volume that can be sent are:
Length (L) = 0 inches
Width (w) = 27 inches
Height (w) = 27 inches
Wondering if you have a name.
2L+2G=108
L=54-G
volume=LG^2=G^2(54-G)
dv/dg=2G*54-3G^2=0
G=108/3=39
L=54-39=15
check my math