HELP PLEASE, can you show steps to, WILL SEND $$$$$!!!!

Question

A coffee company purchases groups of coffee beans and then divides each group of coffee beans into premium beans, regular beans, and unusable beans. The company needs at least 280 tons of premium beans and 200 tons of regular beans. The company may purchase any amount of coffee beans from two suppliers. A group of beans from Supplier A is 20% premium, 50% regular, 30% unusable, and costs $125 per ton. A group of beans from Supplier B is 40% premium, 20% regular, 40% unusable, and costs $200 per ton. How many tons should the company purchase from each supplier to meet its needs at minimum cost?

Answer 1

This is a linear programming problem. You may solve graphically or using a package that performs lp; even Excel does.

Let the pounds of beans from supplier A be a and the pounds of beans from supplier B be b

A beans are $125/ton and B beans are $200/ton, so the cost is 125a + 200b

As beans from A are 20% premium and beans from B are 40% premium, premium beans = .2a + .4b

As beans from A are 50% regular and beans from B are 20% regular, regular beans = .5a + .2b

We need 280 tons of premium beans and 200 tons of regular beans.

Thus, the lp problem is

min 125a + 200b subject to

.2a + .4b >= 280

.5a + .2b >= 200

a, b >= 0

The four constraints intersect at 6 points. At lEast 3 will generally be irrelevant, and we can see at the other points where the objective function is a minimum.

a = 0, b = 0 meet at (0, 0), the first irrelevant point.

.2a + .4b = 280, a = 0 meet at .4b = 280, or b = 700 (0, 700)

.2a + .4b = 280, b = 0 meet at .2a = 280, or a = 1400 (1400, 0)

.5a + .2b = 200, a = 0 meet at .2b = 200, or b = 1000 (0, 1000)

.5a + .2b = 200, b = 0 meet at .5a = 200, or a = 400 (400, 0)

.2a + .4b = 280

.5a + .2b = 200

Double the second equation

a + .4b = 400

Subtract the top equation from the bottom equation

.8a = 120

a = 150

a + .4b = 400

.4b = 250

b = 625

(150, 625)

Note that, with (0, 700) and (0, 1000), (0, 700) does not meet the other constraint.

With (1400, 0) and (400, 0), (400, 0) does not meet the other constraint.

Thus, we have 3 points to consider, (0, 1000), (150, 625), (1400, 0)

Plug each into 125a + 200b

We get $200,000, $143,750, $175,000

The minimum cost is $143,750 with150 tons of A and 625 tons of B

If you wish me to show you how to solve this in Excel, we can set up a session.

Answer 2

Step1. Setup System of Inequality Equation

0.2X+0.4Y>=280

0.5X+0.2Y>=200

It is obvious X>0 and Y>0

Step2. Solve this system, we got

If 0<X<=150. Y>=-0.4X+1000

If X>=150. Y>=-0.5X+700

Step3. Got 3 key points which satisfied the inequality group equation.

They are (X=0, Y=1000), (X=150, Y=625) and (X=1400, Y=0)

Step 4. Check the COST(=X*125+Y*200) for the 3 key points,

which represent how many ton of beans to purchase.

(X=0, Y=1000) COST is 200,000

(X=150, Y=625) COST is 143,750

(X=1400, Y=0) COST is 175,000

Step5. Got the answer.

To purchase supplier A 150 tons and supplier B 625 tons coffee beans can cost minimum

Is it clear and understandable? Message me if you still have any question about it.

You are always welcome.