Question

A plane delivers two types of cargo between two destinations. Each crate of cargo I is 7 cubic feet in volume and 131 pounds in weight, and earns $20 in revenue. Each crate of cargo II is 7 cubic feet in volume and 262 pounds in weight, and earns $25 in revenue. The plane has available at most 525 cubic feet and 12,576 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 1

crates of cargo 2

maximum revenue $

2. National Business Machines manufactures x model A fax machines and y model B fax machines. Each model A costs $100 to make, and each model B costs $150. The profits are $45 for each model A and $30 for each model B fax machine. If the total number of fax machines demanded per month does not exceed 2500 and the company has earmarked no more than $600,000/month for manufacturing costs, how many units of each model should National make each month to maximize its monthly profit?

(x, y)

=

What is the optimal profit?

Answer 1

1.

Itz given plane delivers two types of cargo between two destinations.

Each crate of cargo I

volume=7 cubic feet

weight=131 pounds ,

and revenue earned=$20.

Each crate of cargo Il

volume=7 cubic feet

weight=262 pounds ,

and revenue earned=$25.

plane restrictions

the plane can carry no more than 525 cubic feet of crates ,no more than 12,576 pounds

let x be number of crates of cargo l

let y be the number of crates of cargo ll

Constraint:

The plane can carry no more than 525 cubic feet of crates.

( cubic feet of cargo l) + (cubic feet of cargo 2) ≤ 525 ft³

Constraint:

The truck can carry no more than 12,576 pounds.

( weight of A) + (weight of B) ≤ 12,576lbs

maximizing earnings subject to constraints

maximize:z=20x+25y

subjected to

from equation 1 we can write

x=75-y....(3)

substituting equation (3) in equation (2) we get

(75-y)+2y=96

75+y=96

y=96-75

y=21

x=75-y

x=75-21

x=54

z=20x+25y

z= 54*20+ 21*25

z=$1605

the maximum revenue that can be earned is $1605

2.

Itz given that National Business Machines manufactures x model A fax machines and y model B fax machines.

model A fax machines

cost to make=$100

profit=$45

model B fax machines

cost to make=$150

profit=$30

restrictions

total number of fax machines demanded per month does not exceed 2500.

The company has earmarked no more than $600,000/month for manufacturing costs,

Constraint:

Its given that total number of fax machines demanded per month does not exceed 2500 .

number of  model A fax machines(x)+number of  model B fax machines(y)2500

Constraint:

its given that the company has earmarked no more than $600,000/month for manufacturing costs,

(cost of making model A fax machine*number of  model A fax machines(x))+(cost of making model B fax machine*number of  model B fax machines(y))600,000

Non-negativity: x ≥ 0, y ≥ 0.

maximizing profit subject to constraints

maximize:z=45x+30y

subjected to:

x ≥ 0,

y ≥ 0.

Let u be the number of fax machines less than 2500 made.

Let v be the amount of money not spent from the budget.

fax machines made: x + y + u = 2500

Costs: 100x + 150y + v = 600000

Non-negativity: x ≥ 0, y ≥ 0, u ≥ 0, v ≥ 0.

Objective: Maximize z= 45x + 30y

Put these equations into the following augmented matrix:

The bottom row is the objective function rewritten as: −45x − 30y + z = 0.

Right now x and y are called non-basic variables and u, v, and z are called basic variable.

The corner points of the feasible region correspond to letting the non-basic variables be equal to 0.

Right now, we would let x = 0 and y = 0 which tells us that u = 2500, v = 600000 and z = 0.

This means that if we make zero units of model A and zero units of model B then there are no 2500 fax machines, 600000 unused money from the budget, and 0 profit.

Next we want to switch which variables are basic and non-basic in a way that increases profit. How to choose which columns to switch: Look for the largest negative number in the bottom row of your augmented matrix (if all entries are positive then you are done).

Now we have y and u as the non-basic variables so we have y = 0, x = 2500, u = 0, v = 2,25,000, and z = 1,12,500.

Since we have no negative numbers left in the bottom row, we are done. This means the solution we got above is optimal.

This means the maximum profit occurs when National makes 2500 model A fax machines and 0 model B fax machines. This results in a profit of $1,12,500 and $2,25,000 unspent from the budget.