Respuesta :
Answer:
x = 20 in
L = 40 in
Step-by-step explanation:
Solution:-
- Denote the following:
The side of square cross section = x
The length of package = L
- Given that the combined length "L" of the package and girth "P" of the package must be less than and equal to 120 in
- The girth of the package denotes the Perimeter of cross section i.e square:
P = 4x
- The constraint for our problem in terms of combined length:
L + 4x ≤ 120
L = 120 - 4x .... Eq1
- The volume - "V" -of the rectangular package with a square cross section is given as:
V = L*x^2 ... Eq2
- Substitute Eq1 into Eq2 and form a single variable function of volume "V":
V(x) = 120*x^2 - 4x^3
- We are asked to maximize the Volume - " V(x) " - i.e we are to evaluate the critical value of "x" by setting the first derivative of the Volume function to zero:
d [ V(x) ] / dx = 240x - 12x^2
240x - 12x^2 = 0
x*(240 - 12x) = 0
x = 0, x = 20 in
- We will plug in each critical value of "x" back in function " V(x) ":
V (0) = 0
V(20) = 120(20)^2 - 4(20)^3
= 16,000 in^3
- The maximizing dimension of cross section is x = 20 in, the length of the parcel can be determined by the given constraint Eq1:
L = 120 - 4*20
L = 40 in
- The maximum volume of the rectangular package is with Length L = 40 in and cross section of Ax = ( 20 x 20 ):
The dimension computed shows that the maximum volume will occur at 20.
How to calculate the dimensions?
From the information given, they dimensions can be represented as:
4x + l = 120
l = 120 - 4x
The volume will be:
= x²l
= x²(120 - 4x)
= 120x² - 4x³
We'll use first order differentiation. This will be:
240x - 12x² = 0
12x² = 240x
x = 20
In this case, the maximum volume will occur at 20.
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