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 ):
Answer:
The rate of change is 1 mile per hour per month
Step-by-step explanation:
We are given
initial speed = 3 mph
final speed =6 mph
total number of months =3
now, we can use rate of change formula
we know that
rate of change = ( final speed - initial speed)/(total number of months)
now, we can plug values
and we get
Rate of change is
mph per month
=1 mph per month
The figure is a parallelogram.
Answer:
60 is your answer.
Step-by-step explanation:
x - 8 = 10
add 8 to both sides,
x = 18
substitute 18 for x in the second equation:
4(18) - 12 = 72 - 12 = 60
Please lmk if you have questions.
Answer:40
Step-by-step explanation:
On a fair die, each face has a 1/6 chance of landing on each toss.
5 and 6 are the only faces greater than 4, therefore there is a 1/6 + 1/6 = 2/6 chance of the die landing on a number greater than 4 for each toss.
Multiply the probability by the total number of tosses to get the number of tosses that are greater than 4.
120 * (2/6) = 240/6 = 40
Approximately 40 tosses will land on a number greater than 4.
