```Question 23255
I DO NOT KNOW YOUR BACK GROUND KNOWLEDGE ON THIS TOPIC OF LINEAR PROGRAMMING AND YOUR PRESENT COURSE OF STUDY FOR ME TO EXPLAIN IN REQUIRED DETAIL.FOR THE PRESENT SEE THE FOLLOWING EXAMPLE WHICH IS A LITTLE MORE COMPLICATED AND TRY TO UNDERSTAND THE PROCEDURE AS THIS IS AN IMPORTANT TOPIC WHICH IS BEST DONE BY GOOD UNDERSTANDING ONLY.THEN TRY TO ATTEMPT ANSWERING YOUR PROBLEM.IN CASE OF DIFFICULTY,COME BACK GIVING ME THE ABOVE FEED BACK ON YOUR PRESENT STATUS AND I SHALL ANSWER YOUR PROBLEM IN FULL.

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A mountain climber planning an epedition is concerned about two types of synthetic food. One food contains 100 calories per ounce, 24 units of protein per ounce, and 4 units of fat per ounce. A second food contain 125 calories per ounce, 20 units of protein per ounce, and10 units of fat per ounce. If the man wants a minimum of 2000 calories, 400 units of protein, and 100 units of fat per day, which food or food combination should he use to meet the minimum daily requirements and minimize the total wight?
use two variables and translat the constraints to a system of linear inequalities. Graph the system, and find the vertices, apply linear programming theory and interpret the results.
I SHALL GIVE YOU THE METHOD ,EQNS.,ETC.BUT I WOULD LIKE TO KNOW YOUR TRIALS AND WHERE YOU WERE GETTING STUCK OR GETTING THE WRONG ANSWER.SO PLEASE GIVE A FEED BACK IMMEDIATELY AND ON THAT BASIS ,I SHALL SHOW YOU THE FULL SOLUTION.
LET US DESIGNATE THE 2 FOODS BY F AND G .
1..OUNCE..OF..F..CONTAINS..100..CALORIES..24..PROTEIN UNITS..4..FAT UNITS
1..OUNCE..OF..G..CONTAINS..125..CALORIES..20..PROTEIN UNITS..10.FAT UNITS
MINIMUM REQUIREMENT PER DAY
2000..CALORIES...400...PROTEIN UNITS.....100....FAT UNITS.
LET THE MOUNTAINEER TAKE X OUNCES OF FOOD F AND Y OUNCES OF FOOD G.HENCE HIS TOTAL
CALORIE INTAKE =100X+125Y..THIS SHOULD BE ATLEAST 2000..SO..
100X+125Y>=2000...OR...DIVIDING BY 25 WE GET
4X+5Y>=80............I
PROTEIN INTAKE = 24X+20Y..THIS SHOULD BE ATLEAST 400...SO..
24X+20Y>=400...DIVIDING BY 6 WE GET
6X+5Y>=100...........II
FAT INTAKE =4X+10Y....THIS SHOULD BE ATLEAST 100...SO..
4X+10Y>=100.......DIVIDING BY 2 WE GET ..
2X+5Y>=50.........III
FURTHER OTHER OBVIOUS CONSTRAINTS ARE X>=0 AND Y>=0...WITH THESE CONDTRAINTS WE SHOULD MAKE LEAST WEIGHT OR X+Y SHOULD BE MINIMISED..
SO SAY IF Z=X+Y THEN WE SHOULD FIND MINIMUM VALUE OF Z SUBJECT TO ABOVE CONSTRAINTS..
THIS COMPLETES THE FORMULATION OF EQUATIONS/INEQUALITIES FOR CONSTRAINTS AND THE FUNCTION TO BE OPTIMISED...NOW ON TO SOLUTUION
PLOT ALL THE EQUATIONS ON A GRAPH.X>=0 AND Y>=0 MEANS WE SHOULD CONSIDER ONLY FIRST QUADRANT IN OUR SOLUTION.SO DRAW GRAPHS FOR EQNS I,II AND III IN FIRST QUDRANT ONLY,TAKING EQUALITY.THEN MARK THE ZONES OF INEQUALITIES AS THE REGIONS BEYOND THOSE LINES.YOU GET A POLYGON.FIND THE VERTICES OF POLYGON.AT EACH VERTEX FIND Z=X+Y...FIND WHERE IT IS MINIMUM.HOPE YOU CAN DO IT. REPLY IN CASE OF DIFICULTY.THE GRAPHS WIL LOOK LIKE THIS INCLUDING THE OPTIMISING FUNCTION
Z=X+Y AS EQUAL TO 10,16,50/3,20 and 25 AS A TRIAL.CAN YOU TELL WHY WE HAVE TAKEN THESE FOR TRIAL.ON THIS INFORMATION CAN YOU TELL WHAT IS THE MINIMUM VALUE OF X+Y SATISFYING ALL THE GIVEN CONSTRAINTS?IS IT X+Y=20?PLEASE ANALYSE AND TRY TO UNDERSTAND.IN CASE OF DIFFICULTY COME BACK.

{{{ graph(500,500,0,30,0,30,16-0.8*x,20-1.2*x,10-0.4*x,10-x,16-x,(50/3)-x,20-x,25-x) }}}

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