Tom K. answered • 03/17/21

Knowledgeable and Friendly Math and Statistics Tutor

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.

Lebron J.

Thank You will do!03/18/21