Question 1129285: A farmer can buy two types of plant food, mix A and mix B. Each cubic yard of mix A contains 20 pounds of phosphoric acid, 30 pounds of nitrogen, and 5 pounds of potash. Each cubic yard of mix B contains 10 pounds of phosphoric acid, 30 pounds of nitrogen, and 10 pounds of potash. The minimum monthly requirements are 400 pounds of phosphoric acid, 990 pounds of nitrogen, and 210 pounds of potash. If mix A costs $20 per cubic yard and mix B costs $40 per cubic yard, how many cubic yards of each mix should the farmer blend to meet the minimum monthly requirements at a minimum cost? What is this cost?
Found 2 solutions by ankor@dixie-net.com, ikleyn:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A farmer can buy two types of plant food, mix A and mix B.
Each cubic yard of mix A contains 20 pounds of phosphoric acid, 30 pounds of nitrogen, and 5 pounds of potash.
Each cubic yard of mix B contains 10 pounds of phosphoric acid, 30 pounds of nitrogen, and 10 pounds of potash.
The minimum monthly requirements are 400 pounds of phosphoric acid, 990 pounds of nitrogen, and 210 pounds of potash.
If mix A costs $20 per cubic yard and mix B costs $40 per cubic yard, how many cubic yards of each mix should the farmer blend to meet the minimum monthly requirements at a minimum cost
? What is this cost?
:
a = amt of A mix
b = amt B mix
Write an equation for each ingredient and put in the slope/intercept form
:
Phos Acid
20a + 10b = 400
Simplify divide by 10
b + 2a = 40
b = -2a + 40, red
Nitrogen
30a + 30b = 990
simplify divide by 30
a + b = 33
b = -a + 33, green
Potash
5a + 10b = 210
divide by 10
.5a + b = 21
b = -.5a + 21, blue
;
Plot these three equations
two intersections that interest us, a=7, b=26 and a=24, b=9
Find the cost when a=7, b=26
20(7) + 40(26) = $1180
Find the cost when a=24, b=9
20(24) + 40(9) = $840, this is the obvious choice, A=24 cu/yds, B=9 cu/yds
Answer by ikleyn(52786)  (Show Source):
You can put this solution on YOUR website! .
A farmer can buy two types of plant food, mix A and mix B.
Each cubic yard of mix A contains 20 pounds of phosphoric acid, 30 pounds of nitrogen, and 5 pounds of potash.
Each cubic yard of mix B contains 10 pounds of phosphoric acid, 30 pounds of nitrogen, and 10 pounds of potash.
The minimum monthly requirements are 400 pounds of phosphoric acid, 990 pounds of nitrogen, and 210 pounds of potash.
If mix A costs $20 per cubic yard and mix B costs $40 per cubic yard, how many cubic yards of each mix should the farmer blend
to meet the minimum monthly requirements at a minimum cost? What is this cost?
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Let X be the amount of the mix A (in pounds), and Y be the amount of the mix B.
The constraints are
20X + 10Y >= 400 (1) (phosphoric acid)
30X + 30Y >= 990 (2) (nitrogen)
5X + 10Y >= 210 (3) (potash)
X >= 0, Y >= 10.
The objective function to minimize is F(X,Y) = 20X + 40Y.
In equivalent (and simplified) form the constraints are
2X + 1Y >= 40 (1')
X + Y >= 33 (2')
X + 2Y >= 42 (3')
The feasibility area is the infinite area in the first quadrant which lies ABOVE each of the following straight lines :
- red (which represents the constraint (1'));
- green (which represents the constraint (2')), and
- blue (which represents the constraint (3')).
Plot 2X + 1Y = 40 (constr. (1'), red); X + Y = 33 (constr. (2'), green) and X + 2Y = 42 (constr. (3'), blue)
The four corner points of the feasibility area are
P1 = (0,40) (red line y-intercept);
P2 = (7,26) (red and green lines intersection);
P3 = (24,9) (green and blue lines intersection), and
P4 = (42,0) (blue line x-intercept).
According to the Linear Programming method, we should calculate and compare the values of the objective function at these 4 points.
These values are
P1: F(0,40) = 20*0 + 40*40 = 1600;
P2: F(7,26) = 20*7 + 40*26 = 1180;
P3: F(24,9) = 20*24 + 40*9 = 840;
P4: F(42,0) = 20*42 + 40*0 = 840.
Comparing these numbers, you see that any value of (X,Y) that lies on the segment [P3,P4] satisfies the minimum requirements.
ANSWER. Any value of (X,Y) that lies on the segment [P3,P4] satisfies the minimum requirements. The minimal price value is $840.
Solved.
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