1 MATH 16A LECTURE. OCTOBER 14, 2008. PROFESSOR: SO WELCOME BACK. TESTING, TESTING. I THINK IT WORKS. SO MOVE IT A LITTLE HIGHER. SO OUR GOAL TODAY IS ACTUALLY TO FINISH CHAPTER TWO. ALL THOSE EXAMPLES OF OPTIMIZATION. MAXIMIZING ONE THING OR ANOTHER. SO I'M GOING TO DO A FEW MORE EXAMPLES FROM BUSINESS AND MEDICINE. AND I'M GOING TO THROW IN ONE SMALL VARIATION ON WHAT'S IN THE BOOK, HOW TO, HOW A GOVERNMENT SHOULD MAXIMIZE ITS TAXES. WE HAVE A SAME KIND OF PROBLEM. JUST A SLIGHT VARIATION. SO THAT'S GOING TO BE THE PLAN FOR TODAY. SO LET ME DO A QUICK REVIEW OF THE PROBLEM WE DID LAST TIME BECAUSE I DIDN'T QUITE FINISH IT. AND THAT WAS INVENTORY CONTROL. OR IN PARTICULAR, TO RUNNING A BUSINESS YOU HAVE TO HAVE A CERTAIN AMOUNT OF STUFF TO SELL, YOU WANT TO MINIMIZE THE INVENTORY COSTS. SO I WANT TO JUST QUICKLY REVIEW THAT AND GENERALIZE THAT PROBLEM SLIGHTLY. SO WHAT WERE, WHAT WAS OUR MODEL THERE? WE NEED TO ORDER A CERTAIN SUPPLY OF STUFF EVERY YEAR AND HOW DO WE DECIDE HOW TO DO THAT? WELL, WE'RE GOING TO HAVE TO CHOOSE HOW MANY ORDERS WE MAKE. AND WE GAVE THAT A NAME, R-WAS THE NUMBER OF ORDERS. AND X-WAS THE SIZE OF THE ORDER. TWO VARIABLES THAT WE'RE GOING TO DETERMINE AS I KEEP GIVING YOU DATA. AND SO THIS PARTICULAR PROBLEM YOU NEED A CERTAIN SUPPLY EVERY YEAR. SO R, HOW MUCH IN ONE YEAR HOW MUCH ARE YOU GOING TO ORDER. R-IS NUMBER OF ORDERS PER YEAR. SO IF YOU MULTIPLY THE NUMBER OF ORDERS TIMES THE SIZE OF EACH ORDER, AMOUNT ORDERED PER YEAR. AND FOR THE PARTICULAR PROBLEM WE HAD, 2 WHICH WAS I THINK ORANGE JUICE IT WAS 1200 CASES OF JUICE PER YEAR. THAT WAS THE PARTICULAR EXAMPLE THAT WE HAD DONE. SO R-TIMES X-. THAT WAS ONE PARTICULAR CONSTRAINTS. THE OTHER THING WE WANT TO MINIMIZE IS THE COST. THERE WERE TWO COSTS. ONE WAS HOW MUCH IT COST TO KEEP ALL THOSE SUPPLIES IN A WAREHOUSE, THAT WAS CALLED CARRYING COST. AND THAT DEPENDS ON HOW MUCH IS STORED IN INVENTORY. SO IT DEPENDED ON THE AVERAGE INVENTORY AND THE FORMULA WE USED FOR THAT, SO THE INVENTORY, AS SOON AS YOU ORDER THIS THAT'S YOUR INVENTORY, THAT MEANS YOU HAVE X-INVENTORY IN YOUR WAREHOUSE. SO ON AVERAGE IT WAS GOING TO BE X-OVER TWO THINGS IN THE WAREHOUSE. AND EACH OF THOSE WAS GOING TO COST YOU $8. SO THIS WAS THE CARRYING COST. IT WAS SOME NUMBER, SOME CONSTANTS TIMES X-. X-IS THE SIZE OF THE AMOUNT OF STUFF YOU ORDERED. STUDENT: WHY IS IT DIVIDED BY TWO. PROFESSOR: BECAUSE WHEN WE SAID, LET ME DRAW A PICTURE WE HAD ON THE BOARD LAST TIME OF HOW MUCH WAS IN YOUR INVENTORY. DEPENDING ON TIME, SO ON JANUARY 1ST YOU ORDER X. AND THEN EACH DAY YOU SOLD A LITTLE BIT. SO THE INVENTORY WENT GRADUALLY DOWN TO ZERO. AND YOU WERE CLEVER AND ORDERED ANOTHER BOX OF JUICE TO ARRIVE. SO X-MORE ARRIVED AND THEN YOU USED IT UP. AND THAT'S WHAT YOUR INVENTORY LOOK LIKED DAY BY DAY IN YOUR FACTORY. SOTHE AFFIRM WAS HALF OF THAT, X-OVER TWO. WHAT WE REALLY CARE ABOUT IS SOME CONSTANT TIMES WHICH, WHICH IS HAD A YOU ORDERED. THE THIRD THING WAS ORDERING COST. BECAUSE EVERY TIME YOU MAKE 3 AN ORDER YOU HAVE TO PAY FOR SHIPPING, HANDLING, ALL THAT KINDS OF STUFF. AND THAT WAS SOME NUMBER YOU PAID PER AMOUNT OF STUFF. $75 PER ORDER. IF THERE'S R-ORDERS, THAT'S 75 TIMES RANDOM SAMPLE IS THE COST. IF I PUT THESE TWO TOGETHER, TWO AND THREE, THE TOTAL COST IS GOING TO BE THE SUM, SO THIS IS FOUR X, PLUS 75 R. AND THAT'S WHAT WE WANT TO MINIMIZE. SO WE WANT TO MINIMIZE THE COST. AND WE HAVE A CONSTRAINT WHICH IS R-TIMES X-WAS 1200. SO THAT WAS THE GENERAL SETUP FROM LAST TIME. SO SUMMARIZE, MINIMIZE COST WHICH WAS FOUR X-PLUS 75 R, WITH CONSTRAINT WHICH WAS THAT WE HAD TO ORDER 1200 CASES OF JUICE PER YEAR, TO 1200 OVER X-OF THAT'S WHERE WE HAVE LAST TIME AND THEN WE JUST SOLVED IT. I'M NOT GOING TO DO ALL THE ALGEBRA BUT DO YOU REMEMBER THE GENERAL APPROACH OF HOW WE SOLVE THESE THINGS? THE APPROACH WAS SOLVE CONSTRAINTS FOR ONE OF THE VARIABLES, DOESN'T MATTER WHICH ONE, WHICH ONE DID WE PICK LAST TIME? WAS IT X? SOLVE IT FOR X-. SO THAT'S GOING TO BE 1200 DIVIDED BY R. AND WE SUBSTITUTED THAT INTO THE FUNCTION WE WANT TO MINUTES MAXIMIZE. SO I HAVE A FUNCTION NOW WHICH DEPENDS ON ONE VARIABLE R, AND THEN WE USE CALCULUS TO MINIMIZE THIS THING WHICH WAS 4800 OVER R-PLUS 75 R. THAT'S WHERE WE LEFT IT. AT THAT POINT IT WAS A CALCULUS PROBLEM. SORT OF STRAIGHTFORWARD AND I DON'T WANT TO DO THE ALGEBRA AGAIN. STUDENT: WOULD YOU CALCULATE OF DERIVATIVE OF THAT AND THEN TO FIND THE -- PROFESSOR: SO THAT'S ONE OF OUR, SO THE STANDARD PROCESS KEEP OF 4 DERIVATIVE OF THAT EXPRESSION WITH RESPECT TO R, FIGURE OUT WHERE IS IT IS ZERO AND MAKE SURE IT'S A RELATIVE MINUTE. BECAUSE THE FUNCTION LOOKS LIKE THIS AND I CAN CONFIRM IT BY YES, IT'S CONCAVE UP. SO ALL THOSE TECHNIQUE FROM THE BEGINNING OF THE CHAPTER ARE FAIR GAME. SO WHAT I WANT TO DO NOW IS, SO THIS IS A SPECIAL CASE OF TOTALLY GENERAL PROBLEM IN ECONOMICS. AND I'VE ALMOST SOLVED THE TOTALLY GENERAL PROBLEM HERE. AND I'M GOING DO WRITE IT DOWN. CHOOSE R-AND X-TO MINIMIZE THE COST. AND I WANT TO DO IT WHETHER