Question 1170969: All classes of livestock require the basic nutrients; protein, carbohydrates, lipids, minerals, vitamins, and water. Crude protein is essential to raising healthy livestock and is measured as either a % of the total ration or on an amount per day basis. In this problem, we will use protein hired by Clayton Farms to be in charge of animal nutrition.
The feed orders that you are responsible for have specific dietary requirements that need to be met by combining different ingredients. Because of differences in ingredient prices, you need to determine the least expensive formulation that will meet the nutritional requirements of the feed orders. Clayton Farms has just received an order for 20 tons of feed with 18% crude protein. In this problem, there are two constraints, weight and crude protein.
The available ingredients and their respective prices are given in the table below:
Ingredient
% Crude Protein by Weight
Cost in Dollars per Ton
Barley
11%
$131.81
Corn
9%
$133.63
Cotton Seed Meal
41%
$273.99
Soybean Meal
48%
$301.00
Use the information in the table to find the combination of two ingredients that satisfies the crude protein requirement for the least amount of investment. You will need to convince the owner of Clayton Farms that you have indeed found the least expensive option.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let a = the number of tons of barley
let b = the number of tons of corn
let c = the number of tons of cotton seed meal
let d = the number of tons of soybean meal.
you will be looking at combinations of 2 of these ingredients.
you will use the combination formula of c(n,x) to find the number of groups possible.
you will then analyze each group to see if it's possible for that group to produce the required percentages.
you will then price out the groups that are possible.
you have 4 ingredients.
you will pair these ingredients with each other.
the formula to determine the number of pairings is c(n,x) = n! / (x! * (n-x)!)
with 4 possible pairings, the formula becomes c(4,2) = 4! / (2! * 2!).
this is equal to (4*3*2!) / (2! * 2!) which becomes 4*3/(2*1) which is equal to 6.
the 6 possible pairings are ab, ac, ad, bc, bd, cd.
the ab pairing combines 11% and 9% which will never be equal to 18%, so this pairing is rejected.
the cd pairing combines 41% and 42% which will never be equal to 18%, so this pairing is rejected.
the possible pairings are ac, ad, bc, bd.
these are the pairings that are analyzed for cheapest cost.
since .18 * 20 = 3.6, we'll work with that, rather than .18 * 20.
these pairings can be graphed.
graphing is the quickest way to a solution if you have a good graphing calculator available.
such a calculator can be found at https://www.desmos.com/calculator
i'll graph all pairings, even though the ab and cd pairings are not feasible, just to show you why they're not feasible.
first, the ab pairing is shown below.
in this graph, x represents and y represents b.
second, the ac pairing is shown below.
in this graph, x represents a and y represents c.
third, the ad pairing is shown below.
in this graph, x represents a and y represents d.
fourth, the bc pairing is shown below.
in this graph, x represents b and y represents c.
fifth, the bd pairing is shown below.
in this graph, x represents b and y represents d.
sixth, the cd pairing is shown below.
in this graph, x represents c and y represents d.
you can see that, in the first and sixth pairing, the intersection point has a negative value.
since both values must be positive, the first and sixth pairing do not provide a feasible solution.
the second, third, fourth, and fifth pairing do provide a feasible solution.
these pairings will be analyzed for the least cost solution.
the cost for the second pairing is 3299.75
the cost for the third pairing is 3276.41
the cost for the fourth pairing is 3462.13
the cost for the fifth pairing is 3445.01
the combination with the least cost is the third pairing.
that pairs a with d which is barley with soybean meal.
the cost for the third pairing was derived as follows:
16.216 * 131.81 + 3.784 * 301 = 3276.41
you could have solved these algebraically, rather than by graphing.
in that case you need to solve pairs of equations simultaneously.
i'll do the first pairing to show you how that could be done.
i'll use substitution.
the first pairing has the following two equations that need to be solved simultaneously.
x + y = 20
.11 * x + .09 * y = 3.6
3.6 is the simplification of .18 * 20.
in the first equation, solve for y to get:
y = 20 - x
in the second equation, replace y with 20 - x to get:
.11 * x + .09 * (20 - x) = 3.6
simplify this equation to get:
.11 * x + 1.8 - .09 * x = 3.6
combine like terms and subtract 1.8 from both sides of the equation and simplify to get:
.02 * x = 1.8
solve for x to get:
x = 1.8 / .02 = 90
in the first equation of x + y = 20, replace x with 90 to get:
90 + y = 20
solve for y to get:
y = -70.
your solution is x = 90 and y = -70
since both x and y must be positive, this solution is infeasible.
compare this solution with the graphing solution and you will see that they both provide the same answer, namely:
x = 90 and y = -70.
if you know how to use the graphing calculator, it's much faster than solving manually.
there are also software on the web that can solve the system of equation for you easily as well.
one such can be found at https://www.symbolab.com/solver/system-of-equations-calculator
i used it to solve the same equation i solved manually and graphically, and it provided me with the following solution.
that's the same solution i got manually and graphically.
any questions, let me know.
theo
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