A parcel delivery service will deliver a package only if the length plus the girth (distance around, taken perpendicular to the length) does not exceed 112 inches. Find the maximum volume of a rectangular box with square ends that satisfies the delivery company's requirements.
4 s + L = 112 ... L = 112 - 4 s
v = s^2 * L = s^2 (112 - 4 s) = - 4 s^3 + 112 s
find the local max v
Let the width and height be x inches each, (you said the ends are square)
let the length be y inches
your condition: y + 4x < 112
y < 112 - x
V = x^2 y = x^2(112-x) = 112x^2 - x^3
dV/dx = 224x - 3x^2 = 0 for a max
3x^2 = 224x
x = 224/3 , then y < 112/3
sub into V = x^2 y to get the maximum volume
a similar question...
https://www.jiskha.com/questions/621171/A-parcel-delivery-service-will-deliver-a-package-only-if-the-length-plus-the-girth
To find the maximum volume of a rectangular box with square ends that satisfies the delivery company's requirements, we need to define the variables and set up the problem.
Let's denote the length of the rectangular box as L and the side of the square ends as x. Since the box has square ends, its height and width will both be x.
Given that the length plus the girth (distance around, taken perpendicular to the length) does not exceed 112 inches, we can write the equation for the total distance (length + girth) as:
L + 2x + 2x ≤ 112
Simplifying the equation:
L + 4x ≤ 112
L ≤ 112 - 4x
Next, we need to express the volume of the rectangular box in terms of L and x. The volume of a rectangular box is given by:
V = L * x^2
Now we need to maximize the volume, which means finding the maximum value of V. To do that, we can substitute the inequality L ≤ 112 - 4x into the equation for V:
V = (112 - 4x) * x^2
To find the maximum value of V, we can take the derivative of V with respect to x and set it equal to zero. Differentiating V:
dV/dx = 0 - 4x^2 + 112x
Setting the derivative equal to zero:
0 = -4x^2 + 112x
Simplifying the equation:
4x^2 - 112x = 0
x(4x - 112) = 0
This equation implies that either x = 0 (which is not applicable in this context) or 4x - 112 = 0.
Solving for x:
4x - 112 = 0
4x = 112
x = 28
So, x = 28 satisfies the derivative equation.
Now we can substitute x = 28 back into the volume equation to find the maximum volume:
V = (112 - 4(28)) * (28)^2
V = 56 * 784
V = 43904 cubic inches
Therefore, the maximum volume of the rectangular box with square ends that satisfies the delivery company's requirements is 43904 cubic inches.