A plane delivers two types of cargo between two destinations. Each crate of cargo I is 9 cubic feet in volume and 187 pounds in weight, and earns $30 in revenue. Each crate of cargo II is 9 cubic feet in volume and 374 pounds in weight, and earns $45 in revenue. The plane has available at most 540 cubic feet and 14,212 pounds for the crates. Finally, at least twice the number of crates of I as II must be shipped. Find the number of crates of each cargo to ship in order to maximize revenue. Find the maximum revenue. crates of cargo I crates of cargo II maximum revenue $

So maximum when 46 of I grade and 16 of II grade are produced.

Max revenue = 2100

Step-by-step explanation:

Given that a plane delivers two types of cargo between two destinations

                     Crate I                       Crate II

Volume            9                                  9

Weight           187                               374

Revenue         30                                45

Let X be the no of crate I and y that of crate II

Then

Simplify these equations to get

Solving we get

REvenue = 30x+45y

The feasible region would have corner points as (60,0) or (32,16) or (46,16)

Revenue for (60,0) = 1800

                     (32,16) = 1680

                     (46,16)=2100

So maximum when 46 of I grade and 16 of II grade are produced.

Max revenue = 2100

the correct answer is 5

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet