MATH 16 A-LECTURE. OCTOBER 9, PROFESSOR: WELCOME BACK. HELLO, HELLO, TESTING, TESTING. SO

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1 1 MATH 16 A-LECTURE. OCTOBER 9, PROFESSOR: WELCOME BACK. HELLO, HELLO, TESTING, TESTING. SO WE'RE IN THE MIDDLE OF TALKING ABOUT HOW TO USE CALCULUS TO SOLVE OPTIMIZATION PROBLEMS. MINDING THE MAXIMA OR MINIMA OF THING THIS YOU CARE ABOUT. AND SO WE HAVE A A STANDARD APPROACH LAST TIME. LET ME PUT IT UP AGAIN, REVIEW IT BY DOING ONE EXAMPLE AND STRETCH IT A LITTLE BIT. SO HOW DO YOU SOLVE AN OPTIMIZATION PROBLEM? YOU DRAW A PICTURE. SO YOU KNOW EVERYTHING THAT'S GOING ON AND WHAT ALL THE THING ARE THAT CAN VARY. AND THEN YOU LABEL PARTS OF THE PICTURE WITH NAMES. MIGHT BE THE LENGTH OF SOMETHING, THE WIDTH OF SOMETHING, AREA OF SOMETHING. GIVES NAMES TO ALL THE IMPORTANT QUALITY. SO WRITE DOWN, USING THOSE VARIABLES A FORMULA TO OPTIMIZE. AND OPTIMIZE MEANS EITHER MINIMIZE, SOMETHING LIKE COST, HIS OR HER MAXIMIZE IF IT'S SOMETHING LIKE THE AREA OF A GARDEN. SO YOU WRITE DOWN A FORMULA. AND THEN YOU WRITE DOWN ANOTHER FORMULA WHICH REPRESENTS THE CONSTRAINTS ON YOUR VARIABLES. LIKE YOU CAN'T USE MORE THAN 40 FEET OF FENCE. WE DID AT THAT LAST TIME. YOU CAN'T SPEND MORE THAN SO MUCH MONEY. SO WRITE DOWN THE FORMULA FOR THE CONSTRAINTS. SOLVE THE CONSTRAINTS FOR ONE OF THE VARIABLES IN TERMS OF OTHER. SO SOLVE FOR ONE OF THE VARIABLES. AND THEN YOU, LET ME CALL THIS FORMULA ONE, FORMULA TWO, SOLVE THE CONSTRAINTS, THAT'S FORMULA TWO FOR ONE OF VARIABLES. AND THEN YOU SUBSTITUTE THAT INTO FORMULA ONE TO GET AN EXPRESSION IN ONE VARIABLE. SO NOW I HAVE A FUNCTION, AND EXPRESSION IN ONE 2 VARIABLE AND I KNOW HOW TO OPTIMIZE THAT. FINALLY WE USE CALCULUS TO OPTIMIZE. SO THAT WAS THE GENERAL APPROACH WE USED LAST TIME, IN WORDS. AND LET ME JUST ILLUSTRATE IT AGAIN, GIVING YOU A NEW EXAMPLE. DOES ANYBODY ABOUT ANY QUESTIONS ABOUT THAT GENERAL APPROACH BEFORE I GO USE IT AGAIN? OKAY. SO HERE'S AN EXAMPLE. THE RULE, IT'S WILL I I LOOKED IT UP ON THE WEB, THE LOOK HAD IT WRONG. THE U.S. POST OFFICE RULES SAYS IF YOU WANT TO MAKE A PACKAGE, YOU MUST HAVE A CERTAIN SIZE LIMIT. TO GET A CERTAIN GOOD RATE. AND WHAT IS THE RULE? IT'S THE LENGTH PLUS WHAT THE POST OFFICE CHOOSES TO CALL THE GIRTH WHICH I'LL DEFINE IN A MINUTE, THAT'S TO BE AT MOST 108 INCHES. THE BOOK SAID 84, OBVIOUSLY AN OLD BOOK. AND THE GIRTH IS SORT OF THE PERIMETER GOING AROUND THE NARROW WAY. OKAY. SO HERE'S A TYPICAL EXAMPLE. SO HERE'S A TUBE. AND IT MIGHT BE A BLANK L FROM ONE ENDS OF THE TUBE TO THE OTHER. THE THE TUBE HAS RADIUS R, THEN THE GIRTH IS JUST THE SISH CUNCHES OF THE CIRCLE. CIRCUMFERENCE OF THE CIRCLE. SO THAT'S THE, THE GIRTH IS JUST THE SISH COVERAGE OF THE CIRCLE. TWO PI R. ALL THE WAY AROUND. SO THAT'S THE KIND OF SHAPE THAT YOU'RE ALLOWED TO MAIL. SO THE QUESTION IS WHAT SHAPE CYLINDER, HAS THE LARGEST VOWEL. WHAT IS THE MOST AMOUNT OF STUFF THAT YOU CAN MAIL IN THE POST OFFICE WITH THESE RULES. HOW BIG CAN I MAKE THE VOLUME. I'M ALLOWED TO VARY L AND ALLOWED TO VARY R. SO LET'S WRITE DOWN THE CONSTRAINTS IN THE FORMULA TO OPTIMIZE. SO WHAT I WANT TO DO IS MAXIMIZE THE VOLUME. I'LL GIVE THAT A NAME, V SO. IF I KNOW THE 3 RADIUS OF A CYLINDERS AND THE LEARNLINGS OF A CYLINDERS WHAT'S THE VOLUME. THE LENGTH TIMES THE AREA OF THE CROSS SECTION. SO

2 THAT'S GOING TO BE THE LENGTH. AND WHAT IS THE AREA OF THE CROSS SECTION, SO IT'S A CIRCLE OF RADIUS R. SO PI R-SQUARED. OKAY. SO THERE'S THE VOLUME. AND THE CONSTRAINT SAYS THAT THE LENGTH PLUS THE GIRTH, IS TWO PI R, THE SIR CURCHES OF THE CIRCLE IS ONE OWE EIGHT. THAT'S THE FIRST THREE STEPS UP THERE. EVERYBODY BY THAT MODEL? SO I HAVE TWO VARIABLES L AND R. HERE'S THE CONSTRAINT AND I WANT IS IT PICK THEM TO MAXIMIZE THAT. STUDENT: CAN YOU IT BE LESS THAN ONE OWE EIGHT. PROFESSOR: I'M MAKING A LEAP HERE. I'M SAYING THAT DO YOU THINK THE VOLUME IS GOING TO BE THE BIGGEST WHEN I MAKE THIS AS BIG AS POSSIBLE? PROBABLY. SO EVEN THOUGH THIS IS WHAT THE POST OFFICE SAYS, I'M USING YOUR INTUITION TO SAY I MIGHT AS WELL MAKE IT AS BIG AS POSSIBLE, THAT WILL MAXIMIZE THE VOLUME. SO LET'S GO AHEAD AND DO THAT. SO I'M GOING TO SOLVE THE CONSTRAINT. STUDENT: IS THE CONSTRAINT ALWAYS GIVEN. PROFESSOR: IN THE PROBLEM WE'RE TALKING ABOUT YOU CON GET STARTED UNLESS YOU HAVE SOME CONSTRAINT LIKE THIS. HERE IT CAME FROM THE POST OFFICE IT. DEPENDS ON THE SITUATION YOU'RE IN. WE'LL DO OTHER EXAMPLES. SO LET'S FOLLOW THE CONSTRAINT FOR ONE OF THE VARIABLES. AND LET ME MAKE SURE I PICKED THE ONE I LIKED. SO PRETTY EASE TO TO SOLVE. HERE'S THE LENGTH. SUBSTITUTE THAT INTO C. OKAY. AND IF I MULTIPLY THAT OUT I GET 108 PI R-SQUARED 4 MINUS TWO PI SQUARED R-CUBED. THERE'S A FUNCTION OF EXACTLY ONE VARIABLE R-AND I WANT TO MAXIMIZE. SO NOW FINALLY EVERYTHING HAS BEEN ALGEBRA OR A TINY OF GEOMETRY SO FAR. NOW IT'S TIME TO DO CALCULUS. SO, WE CAN PLOT THIS. THAT'S A GOOD IDEA IT CHECK BUT LET ME DO IT THE WAY WE'VE LEARNED SO FAR AM I'M GOING TO SOLVE V PRIME EQUALS ZERO, LOOK FOR THE CRITICAL POINTS WHERE THE TANGENT IS HORIZONTAL. SO LET ME SOLVE IT. SO TWO TIMES 108, THERE'S A DERIVATIVE OF THE FIRST TERM. OKAY. EVERYBODY BUY THAT? THAT'S WHAT THE DERIVATIVE IS. JUST DIFFERENTIATE THE ONE AT A TIME. I LET ME FACTOR OUT. THERE'S A COMMON FACTOR OF R-IN BOTH OF THEM. FACTOR OUT R-. FACTOR OUT PI. AND I'M LEFT WITH 216 MINUS SIX PI R. OKAY. SO I NEED TO SOLVE THAT EQUAL TO ZERO. I FACTORED IT SO THAT'S EQUAL TO ZERO, IF R-EQUALS ZERO BECAUSE OF THAT FACTOR OR 216 EQUAL SIX PI R-AND IF I DIVIDE BY SIX PI I GET R-EQUALS 216 OVER SIX PI WHICH IS 36 OVER PI. SO JUST SIMPLE ALGEBRA. SO NOW I HAVE TWO CRITICAL POINTS. ZERO AND 36 OVER PI. I HAVE TO DECIDE WHICH ONE IS THE MINIMUM AND WHICH ONE IS THE MAX. WE DON'T HAVE TO WORK TOO HARD AM GO UP TO V AND PLUG IN R-EQUALS ZERO. SO INTUITIVELY, TAKING THE RADIUS OF THE TWO TO THE ZERO, WHAT'S THE VOLUME.? ZERO THAT'S A PRETTY MALL TUBE. SO R-EQUALS ZERO IS THE MINIMUM. BECAUSE IMPLIES VOLUME EQUAL ZERO LET ME MAKE SURE THAT THIS OTHER VALUE IS AT RELATIVELY MAXIMUM. CHECK THE SECOND DERIVATIVE TO DO THAT. V DOUBLE PRIME. TAKE THIS ONE AND DIFFERENTIATE IT AGAIN AND I GET 216 PI MINUS 12 PI SQUARED R. AND I HAVE TO ASK IS 5 THAT POSITIVE OR NEGATIVE AT MY POINT WHERE THE TANGENT IS FLAT. SO LET MET PLUG IN V DOUBLE PRIME TIMES 36 OVER PI, IS MINUS 216 PI SO IT'S NEGATIVE. SO THE SECONDS DERIVATIVE IS NEGATIVE. THAT MEANS WHAT IS THE SHAPE? CONCAVE DOWN. SO IT LOOKS LIKE THAT. THAT MEANS IT'S A RELATIVE MAX MA. THAT'S POINTS I WANT TO DEFINE. IF I PLOT THE WHOLE FUNCTION, SO AT ZERO, IT'S ZERO.

3 AND THAT'S WHAT THE WHOLE FUNCTION LOOKS LIKE. AND HERE I HAVE 36 OVER PI. AND SO THE ONLY THING I'VE DONE IS CHECKED THAT THAT POINT, CHECK THE POINT R-EQUALS ZERO AND MADE SURE THAT'S THE MINIMUM. AND THAT'S A MAXIMUM. I DIDN'T HAVE TO PLOT THE ENTIRE GRAPH. IF I PLUG THAT INTO THE FORMULA I WILL GET 36 CUBED OVER PI. THAT IS WHAT HAPPENS WHEN YOU PLUG IN V OF, SO, HOW DID I GET THAT? FILL EVERYTHING ELSE IN. I FIGURED OUT WHAT R-IS, IT'S 36 OVER PI, SO NOW I HAVE TO TELL YOU WHAT L IS, IT'S 108 MINUS TWO PI R. R-IS 36 OVER PI. AND SO 108 MINUS 72. AND THAT'S 36. SO I HAVE L EQUALS 36. R-EQUALS 36 OVER PI AND SO HERE IS THE MAXIMUM VOLUME TO GET, IT'S 36 CUBED DIVIDED BY (ON BOARD) \{^}MENT\{^}.{~}{~}{- } 36 CUBED OVER PI. JUST IN CASE YOU'RE WONDERING, 14,851 CUBIC INK ROUGHLY. SO THERE IS THINK WALKED THROUGH THIS WHOLE PROCESS I PUT UP THERE. THE ONLY TIME I USED CALCULUS IS AT THE ENDS AND USE THE ALL THE STUFF OF MINIMIZING AND MAXIMUMING. MAKE SURE IT'S CONCAVE UP AND CONCAVE DOWN AS APPROPRIATE. ANY QUESTIONS ABOUT THAT EXAMPLE? STUDENT: HOW DID YOU GET R-AND -- 6 PROFESSOR: SO I, DID YOU GET HOW YOU SOLVED FOR THE LENGTH L. SO THEN HE SUBSTITUTE THAT IN THE EXPRESSION FOR THE VOLUME. I GET THE VOLUME IS SOMETHING THAT ONLY DEPENDS ON R. 108 MINUS TWO PI R. SO FAR SO GOOD? SO NOW I KNOW TO MAXIMIZE THAT AS A FUNCTION OF R-AND THAT'S A CALCULUS PROBLEM. DIFFERENTIATE V WITH RESPECT TO R. THAT TELLS ME WHERE THE RELATIVE MINIMA AND MAXIMA COULD BE. AND I JUST HAVE TO TELL -- R-IS 36 OVER PI. AND WHEN R-ZERO THE VOLUME IS ZERO THAT'S NOT THE MAXIMUM. THE OTHER ONE IS THE ANSWER. AND SO R-IS 36 OVER PI AND I DOUBLING CHECKED BY MAKING SURE THE FUNCTION IS CONCAVE DOWN. SO THAT'S THE MAXIMA. AND THE REST IS ARITHMETIC RIGHT. STUDENT: YOU SAID R-TIMES PI EQUAL ZERO AND 16 MINUS. PROFESSOR: I'M SAILING THIS THICK HAS TO BE ZERO THAT MEANS EITHER THAT FACTOR IS ZERO OR THAT FACTOR IS ZERO. JUST EASIER TO DO THAN, I COULD HAVE SAID SOLVE THE QUADRATIC. BUT IT'S JUST EASIER... OKAY. SO I WANT TO DO, IF THERE ARE NO MORE QUESTIONS ABOUT THIS EXAMPLE I WANT TO DO A SLIGHTLY HARDER ONE. PROFESSOR: V DOUBLE PRIME CAME FROM DIFFERENTIATING V PRIME AND THEN I WANT IT KNOW THE PARTICULAR VALUE OF R-THAT I CARE ABOUT WHICH IS 36 OVER PI CONCAVE UP OR DOWN, PLUG IN 306 OVER PI THAT THAT NUMBER AND I GET A NEGATIVE NUMBER SO THAT MEANS IT'S CONCAVE DOWN. STUDENT: DID YOU CYLINDER PART OF THE EQUATION OR ARE WE 7 SUPPOSED TO COME UP WITH THAT. PROFESSOR: THE POST OFFICE SAYS, THE STATEMENT OF THE PROBLEM WAS WHAT'S THE BIGGEST CYLINDER. WE CAN ALSO ASK WHAT GOES THE BILINGSEST, RECTANGULAR BOX. THAT'S ANOTHER INTERESTING QUESTION BUT THAT WOULD COME UP WITH A DIFFERENT ANSWER. LET'S TRY, ACTUALLY LET'S TRY THAT ONE. I THINK THAT'S MY NEXT ONE. AND THIS ONE'S GOING TO BE TRICKIER, IT'S ALMOST AM SAME QUESTION. WHAT IS THE LARGEST VOLUME NOT A CYLINDER OF A RECTANGULAR BOX, SUBJECT TO THE SAME RULE WHICH IS THAT THE LENGTH PLUS THE GIRTH IS LESS THAN OR EQUAL TO 108. I MIGHT ADDS

4 WELL MAKE IT AS BIGGER AS POSSIBLE. KEEP 108 BECAUSE THAT WILL MAKE THE VOLUME AS BIG AS POSSIBLE. SO ME ME START AGAIN. DRAW A PICTURE. SO THERE'S MY RECTANGULAR BOX. TO DESCRIBE THE SHAPE OF THIS I NEED THREE, SO IT WILL BE X-FROM THERE TO THERE, GOING TO BE Y-VERTICALLY, IT'S GOING TO BE Z-AT THREE VARIABLES AM IF I KNOW ALL THREE I KNOW THE SHAPE OF BOX. SO I'LL USE THE SAME NAME V, SO WHAT IS THE SOLVE OF MY BOX? IF I TELL YOU ALL THREES SIDES WHAT IS THE FORMULA FOR THE VOLUME OF A BOX. X-TIMES Y-TIMES Z. NOW I ALSO HAVE TO HAVE THE LENGTH PLUS THE GIRTH. LENGTH PLUS GIRTH. THE LENGTH IS X, AND WHAT IS THE GIRTH? IT'S THE PERIMETER AROUND THE OTHER WAY, SO IT'S THE SUM OF THOSE, THAT IS WHAT IT MEANS. WELL, SO IT'S Z-IN THAT DIRECTION AND Y-IN THE VERTICAL DIRECTION. SO WHAT IS THE PERIMETER AROUND THAT? TWO Y-PLUS TWO Z. BECAUSE IT'S Z-PLUS Y-PLUS Z-PLUS Y. SO THAT'S GOING TO BE LENGTH PLUS GIRTH. AT THIS POINT YOU MIGHT 8 SAY WHOOPS I HAVE THREE VARIABLES. OKAY. LET ME SHOW YOU. SO WHAT WE'RE GOING TO DO IS PICK ONE VARIABLE. I'M PICK X. AND FIXED. FRIEND IT'S LIKE A CONSTANT. SO FIX THE LENGTH OF BOX. AND THEN HOW MANY VARIABLE DO I HAVE LEFT? I HAVE TWO, Y-AND Z. SOLVE THE PROBLEM USING Y-AND Z. WHEN I'M DONE AND SOLVED THE PROBLEM MY ANSWER IS GOING TO BE A FORMULA DEPENDING ON X. THE ANSWER WILL BE THE LARGEST POSSIBLE VOLUME V, WHICH IS GOING TO BE A FORMULA THAT'S GOING TO DEPENDS ON X. X-IS JUST GOING TO BE CARRIED ALONG THROUGH. GOING TO DEPENDS ON X. Y-AND Z-IS ARE GONE, I'VE SOLVE THAT PART OF PROBLEM BUT I'M LEFT WITH SOMETHING THAT DEPENDS ON ONE VARIABLE. AND I WANT TO MAXIMIZE IT. AND SO LET ME DO THAT OVER HERE. THE LAST TIME IS TAKE MY FUNCTION V OF X-WHICH ONLY DEPENDS ON ONE VARIABLE NOW. THE LAST STEP IS FINALLY MAXIMIZE V OF X-WHICH ONLY DEPENDS ON ONE VARIABLE USING CALCULUS. NO CONSTRAINTS. OKAY. SO THIS IS JUST ONE STEP HARDER AM I'M GOING IT ILLUSTRATE THIS. IT MAKE MORE SENSE ONCE I DO AN EXAMPLE. STUDENT: HOW FIGURE OUT THE CONSTRAINT WAS X-PLUS TWO X-DOES. PROFESSOR: THE POST OFFICE SAYS LENGTH PLUS GIRTH. THE GIRTH IS THE PERIMETER THE OTHER WAY. SO IT IS THE PERIMETER EVERYONE X-OF OF THAT RECTANGLE. AND THAT SIDE IS THE TOP AND BOTTOM AT LENGTH Z-AND SIDES OF LENGTH Y. SO TWO Y-PLUS TWO Z. SO FROM NOW I'M -- YES. STUDENT: WAIT WHAT IS THE PERIMETER WON'T THAT BE TWO Y-TIMES Z-PLUS -- SORRY,. 9 PROFESSOR: IT'S OKAY. PERIMETER IS JUST THE LENGTH THAT WE'RE GOING AROUND. IS THAT OKAY. STUDENT: YEAH. PROFESSOR: OKAY. SO LET ME JUST DO IT. HERE'S GOING TO BE THE CONSTRAINT. AND I'M GOING TO PRETENDS AS I DID THAT X-IS CONSTANT. I'M NOT, KEEP IT ALONG AS A CONSTANT. SO IS THE CONSTRAINT SAYS AT THAT TO Y-PLUS TWO Z-EQUALS 108 MINUS X. SO THAT'S JUST A NUMBER. AND THE FORMULA I WANT TO MAXIMIZE IS THE VOLUME AND THAT'S X-TIME Y-TIMES Z-. BUT X-IS NOT JUST LIKE ICON STANLT. IT'S FIXED. THE ONLY TWO VARIABLE IN YOUR MINE NOW SHOULD BE Y-AND Z. SO I HAVE CONSTRAINT IS Y-AND Z-AND A FUNCTION, IT'S EXACTLY THE SAME PROBLEM I HAD BEFORE. X-IS JUST GOING TO GET CARRIED ALONG. SO LET ME SOLVE THE CONSTRAINT, SO REPEAT THAT, X, TREAT X-LIKE A CONSTANT. CARRY IT ALONG. SO

5 NOW LET ME DO THE THING UP THERE. SOLVE THE CONSTRAINT FOR ONE OF MY TWO VARIABLES. Y-OR Z. SO MET SO WILL IT FOR Y, IT'S ABOUT THE SAME EITHER WAY. SO Z-OVER THERE. AND DIVIDE BY TWO. AND I GET 54 MINUS X-OVER TWO MINUS Z. ALL RIGHT. SO I HAVE SOLVED FOR ONE OF THE VARIABLES. AND SO NOW I SUBSTITUTE INTO V. SO X-TIMES Y-TIMES Z-SO THERE'S X-TIMES 54 MINUS X-OVER TWO MINUS Z-TIMES Z. SO NOW WHEN I THINK OF X-CAUSE SON STAND THERE'S JUST ONE VARIABLE IF THIS EXPRESSION CALL Z. SO I CAN ASK CAN YOU PLEASE MAXIMIZE THAT. SO NOW WHAT I'M GOING TO DO IS FIND MAXIMUM WITH RESPECT TO Z. X-IS FIXED. OTHER ALLOWED TO VARY Z. ARE WE OKAY WITH THIS? WHAT'S CONSTANT, X-AND THE OTHER TWO 10 VARIABLES. STUDENT: HOW DID YOU KEEP X-CON DISTANT. PROFESSOR: I COULD HAVE CHOSEN ANY ONE OF THEM, IT'S ARBITRARY AM A GOOD WAY IT CHECK YOUR ANSWER IS PICK ANOTHER ONE AND MAKE SURE YOU GET THE SAME ANSWER. THAT'S A GENERAL GOOD APPROACH. SOLVE TWO DIFFERENT WAYS YOU SHOULD GET THE SAME ANSWER. SO LET ME SIMPLIFY THIS. THIS IS GOING TO BE 504 X-Z. FIFTY-FOUR X-IS JUST CONSTANT. MINUS X-SQUARED OVER TWO TIMES Z, MINUS X-Z-SQUARED. (ON BOARD). SO EVERYTHING IN PREEN THESE IS BASICALLY LIKE A CONSTANT. SO WHAT DO I DO? I DIFFERENTIATE MY FUNCTION VOLUME, AND I'M GOING TO USE THIS SYMBOL IT MAKE SURE I REMEMBER WHAT I'M DIFFERENTIATING, DIFFERENTIATE THAT WITH RESPECT TO Z. SO WHAT THIS IS TERM W TO Z-. 54 X. AND HOW ABOUT THE DERIVATIVE OF THE SECOND TERM WITH RESPECT TO Z. MINUS X-SQUARED OVER TWO. FINALLY THE DERIVATIVE OF THIS TERM WITH RESPECT TO Z-IS MINUS TWO X-Z. (ON BOARD). THAT'S JUST TREAT IT LIKE A CONSTANT. AND SO WHAT I NEED TO DO IS SET THIS, SO NOW WHAT I NEED TO DO IS SOLVE THIS THING D-V D-Z-EQUALS ZERO FOR THE THING I'M ALLOWED TO VARY. Z- STUDENT: WHY DOES Z-STAY WITH THE MINUS TWO X-Z- PROFESSOR: SO WHEN I DIFFERENTIATE THIS IT'S A CONSTANT TIENLZ Z-SQUARED. SO THE DRIVEN IS THE SAME CONSTRAINT TIMES TWO Z. STUDENT: OH, OKAY. PROFESSOR: IF I DIFFERENTIATE THE FIVE Z-SQUARED IT WOULD BE FIVE TIMES TWO Z, OR TEN Z. FIVE, HERE I HAVE A NUMBER FIVE, 11 X-IS JUST LIKE A CONSTANT LIKE FIVE. SO IT'S TWO TIMES THAT CONSTANT TIMES Z. TWO X-Z. SO I NEED TO SOLVE THIS THING EQUAL TO ZERO. SO IT'S, I HAVE 54 X-MINUS X-SQUARED OVER TWO EQUALS TWO X-Z. THAT'S EASY TO SOLVE FOR Z. DIVIDE BY TWO X. SO I GET Z-EQUALS 27 MINUS X-OVER FOUR. (ON BOARD). I HAVE MY ANSWER FOR A PARTICULAR, X-IS AFFIX LENGTH, AND THIS IS WHAT THE OPTIMUM SIZE IS. IN THAT DIRECTION FOR THE WIDTH OF MY BOX. SO STRAIGHTFORWARD CALCULUS. SO I'VE GOTTEN Z. IS THIS OKAY? STUDENT: WHY DO EQUAL TO -- PROFESSOR: SO I'M TRYING IT MAXIMIZE THIS. SO I FIND THE CRITICAL POINT. YOU'RE SAYING I SHOULD DOUBLE CHECK IT MAKE SURE IT'S A MAX AND NOT A MINIMUM. SO LET MET DUB CHECK AND MAKE SURE THE SECOND DERIVATIVE IS WHAT? WHAT SHOULD THE SECOND DERIVATIVE BE? WE WANT IT TO LOOK LIKE THAT SO SHOULD BE NEGATIVE. SO LET ME DOUBLE CHECK. IT MEANS I HAVE TO DIFFERENTIATE AGAIN WITH RESPECT TO Z. SO WHAT IS THE DERIVATIVE OF THIS TERM WITH RESPECT TO Z? ZERO. X-IS A SCON STAND. SO THIS IS JUST A CONSTANT. DERIVATIVE IS CONSTANT. SO IT'S ZERO.

6 SO ZERO HOW ABOUT THIS TERM? ZERO. SO THOSE TERM DON'T MATTER. ALL I GET HERE IS IS WHAT THE THE DERIVATIVE OF THIS WITH RESPECT IT Z? MINUS TWO X-AND THAT'S NEGATIVE. X-IS THE LENGTH. SO IT'S CONCAVE DOWN SO I HAVE INDEED FOUND THE MAXIMUM. STUDENT: EXPLAIN WHAT DID YOU. PROFESSOR: SO I SET THIS TO ZERO. MOVED TWO X-Z-TO THE OTHER SIDE. I GOT AT THAT EQUATION. I WANT TO SOLVE FOR Z-SO DIED 12 BOTH SIDE BY X- STUDENT: (INAUDIBLE). PROFESSOR: TO GO IT HERE TO HERE, 54 X-DIVIDED BY TWO X-IS IS 20 SEX. X-SQUARED OVER TWO -- I'M NOT QUITE DONE. I INSIDE TO FIND OUT WHY ALSO. SO I HAVE Z. I NEED Y. SO WHAT IS MY CONSTRAINT? NOW GOING TO SOLVE THE CONSTRAINT FOR Y. AND IS IT ALREADY WRITTEN DOWN HERE? ANYWAY, SO WHAT DO I GET? I THOUGHT I WROTE THIS DOWN ALREADY. SO LET ME SOLVE FOR Y. SO Y-IS 54 MINUS X-OVER TWO MINUS Z-OKAY. SO WHAT IS 54 MINUS 27? A LITTLE PRACTICE HERE? TWENTY-SEVEN. AND MINUS X-OVER TWO PLUS X-OVER FOUR IS, DOES THAT LOOK FAMILIAR? Y-AND Z-ARE EQUAL. SO THE OPTIMUM SOLUTION TURNS OUT TO BE WHEN Y-AND Z-ARE EQUAL. SO WHAT'S THE SHAPE OF THE CROSS SECTION? IT'S A SQUARE. SO THE OPTIMUM, THE LARGEST VOLUME HAPPENS WHEN Y-EQUALS Z, IN OTHER WORDS I-E- THE CROSS SECTION IS SQUARE. SO THAT'S A NICE LITTLE GEOMETRIC INSIGHT. BUT WE'RE IN THE DONE. YOU KNOW THE ANSWER IS GOING TO BE SQUARED. SO HOW FAR DOWN HAVE I GET END DOWN HERE. PROFESSOR: DOWN TO THE LINE AND SOLVED CONSTRAINT FOR ONE VARIABLE. SO I SOLVED THE PROBLEM. SO I NOW DECIDED THAT THE SHAPE OF MY BOX. IF I CHOOSE THIS SIDE D-X-THEN Y-THAT'S TO BE 27 MINUS X-OUR FOUR AND Z-HAS TO BE 20S SEVEN MINUS X-OVER FOUR. SO NOW I HAVE A BOX, THIS SIDE IS X-. THAT SIDE IS 27 MINUS X-OVER FOUR. THAT SIDE -- THE QUESTION IS HOW DO I GET X-IT MAXIMIZE THE VOLUME. THERE'S ONLY ONE VARIABLE. SO LET ME DRAW 13 THAT PICTURE AGAIN. SO NOW WE KNOW THE LARGEST VOLUME OCCURS WHEN THE BOX LOOK LIKE THE FOLLOWING. IT'S GOING TO BE X-ON THIS, X-ON THAT SIDE. THIS SIDE IS GOING TO BE BETWEEN SEVEN MINUS X-OVER FOUR AND THAT'S SIDE IS GOING TO BE 27 MINUS X-OVER FOUR. A ONE VARIABLE. SO WHAT IS THE VOLUME OF THIS BOX? IT'S MULTIPLY ALL THREE TOGETHER. SO I GET X-TIMES 27 MINUS X-OVER FOUR TIMES 27 MINUS X-OVER FOUR. THAT'S THE VOLUME OF THIS THING. THOSE TWO ARE EQUAL SO MULTIPLY THOSE THREE THINGS TOGETHER AND I GET VOLUME. SO WHAT I NEED TO DO NOW IS MAXIMIZE THIS. NOW WE HAVE ONE VARIABLE. THAT'S AN EASY CALCULUS PROBLEM. I HAVE A FUNCTION OF ONE VARIABLE X. JUST TO KEEP IT CRYSTAL CLEAR, LET ME MULTIPLY THIS OUT. SO I'M GOING TO GET 27 SQUARED TIME X. MINUS 27 OVER TWO X-SQUARED PLUS X-CUBED DIVIDED BY 16. THAT'S WHAT HAPPENS WHEN YOU MULTIPLY IT. EASIER IT MULTIPLY THOSE TWO TOGETHER. SO I WANT TO FIND THE MAXIMUM. SO I DO THE USUAL TRICK. DIFFERENTIATE IT WITH RESPECT IT X. MAKE IT VERY CLEAR, SO X-IS THE ONLY VARIABLE AROUND AT THIS POINT SO DO D-V D-X. AND THAT'S GOING TO BE 27 SQUARED MINUS 27 X-PLUS THREE X-SQUARED DIVIDED BY 16. AND I HAVE TO SET THAT EQUAL TO ZERO. SOLVE FOR D-V D-X-EQUALS ZERO. THAT'S GOING TO GIVE ME THE CRITICAL POINT. AND SO I'M NOT GOING, SO NOW WE NEED, I'M

7 GOING TO SAY USE THE QUADRATIC FORMULA AND THE ANSWER COMES OUT NICE AND NEAT. AND THERE ARE TWO ANSWERS, YOU EITHER GET X-EQUALS 108 OR X-EQUALS 36. I WON'T DO ALL THE ALGEBRA. THE 14 QUADRATIC FORMULA BUT THAT'S WHAT YOU'VE GOT TO DO. SO THERE ARE TWO MINIMUM FOR WHERE THE DERIVATIVE IS ZERO. SO LET'S JUST USE OUR INTUITION. WHAT DO YOU THINK, WHAT'S THE VOLUME WHEN X-IS 108? ZERO. BECAUSE THAT WILL BE ZERO AND THAT WILL BE ZERO AND IT WILL BE A VERY TINY BOX. THIS IS THE MINIMUM. X-EQUALS 108. THE OTHER ONE, IS THE MAXIMUM. LET JUST DOUBLE CHECK THOUGH. TAKE THE SECOND DERIVATIVE AND YOU GET MINUS 27 PLUS SIX DIVIDED BY 16 X-AND YOU CAN DOUBLE CHECK, BUT THAT IS INDEED NEGATIVE. AT X-EQUALS 36. SO THIS, SO WITH A DO WE HAVE? WE HAVE X-EQUALS 36. WHAT'S THE OTHER DIMENSION THEN, 27 MINUS THIRPT SIX OVER FOUR, WHAT IS THE SHAPE? X-DIVIDED BY FOUR IS NINE, 20S -- THAT SIDE'S 18, SO IT'S 36 BY 18 BY 18, THAT IS THE BIGGEST BOX YOU COULD MAKE. STUDENT: WOULD IT WORK TO USE THAT VOLUME EQUATION UP THERE AND SUBSTITUTE IN FOR Z- PROFESSOR: WHICH VOLUME EQUATION. STUDENT: THE ONE AT THE VERY TOP. PROFESSOR: WELL, THE ONLY VOLUME EQUATION IS X-TIMES Y-TIMES Z. HOWEVER YOU WANT IT AOE LESS THAN OR EQUAL TO NAME THE VARIABLES YOU WOULD HAVE GOTTEN THE SAME ANSWER. SO I SAY PRETENDS K-IS CONSTANT. I COULD HAVE PRETENDED Y-IS CONSTANT AND SOLVE FOR -- ANYWAY YOU DID IT YOU SHOULD HAVE GOTTEN THE SAME ANSWER. STUDENT: COMPLAIN HOW YOU KNOW THAT ONE IS THE MINIMUM. PROFESSOR: WHEN X-IS 108. I COULD PLOT IT BUT I'M GOING IT SHORT CIRCUIT THAT. IT'S NOT SO HARD AM IN SIDE IS 108 THEN THIS 15 IS 27 MINUS -- THAT'S ZERO. THE VOLUME OF THE BOX IS THEN ZERO. THAT'S THE SMALLEST BOX YOU COULD HAVE. ZERO VECTOR. STUDENT: (INAUDIBLE). PROFESSOR: THE CONSTRAINT, STARTED UP THERE IT SAID THAT THE LENGTH PLUS THE GIRTH IS 108 INCHES. STUDENT: OKAY. PROFESSOR: AND SO YOU CAN DUCK CHECK OF THE 36 PLUS FOUR TIME 18 IS 108. SO WE SATISFIED THAT CONSTRAINT. SO THIS IS, WE STARTED BY DOING JUST TWO VARIABLES. HERE WE HAVE THEY BY WITH YOU SAID E-MAKE IT TWO BY TREATING ONE VARIABLE AND LEAVING IT OUT. TREATING IT AS A CONSTANT. SO YOU CAN DO INTERESTING PROBLEM BASED ON WHAT YOU KNOW AM THAT WAS THE PURPOSE OF THAT. AND YOU HAD IT MAKE CHOICES. I CHOSE TO MAKE X-CONSTANT. I COULD HAVE CHOSEN Y-OR Z-AND GOTTEN THE RIGHT ANSWER. THERE'S NOT UNIQUE WAY TO GET TO THE RIGHT ANSWER. LET ME GO BACK AND DO A SLIGHTLY SIMPLER PROBLEM AM I THINK THESE EXAMPLES ARE COMPLICATED ENOUGH. THEY DESERVE A FEW MORE ILLUSTRATIONS. STUDENT: YOU GOT THE 108 FROM THE FIRST DERIVATIVE,. PROFESSOR: THIS CAME FROM JUST, QUALITY FORMULA I MULL IT OUT AND DIFFERENTIATE IT. NOW I SET IT EQUAL TO ZERO IT'S A QATDZ RAT I CAN IF YOU GO OF X-. SO I USE THAT, ONE OF THEM GIVES THE ANSWER AND ARE OF 108. THE OTHER, 36. SO I HAVE TO DECIDE WHICH IS MINIMUM AND WHICH IS MAX. WHEN THIS IS 108 THE VOLUME IS ZERO. SO THAT'S THE MINIMUM. 16

8 STUDENT: SO WHY DID YOU TAKE THE SECOND DERIVATIVE. PROFESSOR: JUST, I JUST WANTED TO DOUBLE CHECK IS X-EQUALS IT 36 REALLY MAXIMUM. I WAS DOING EXTRA DUE DILIGENCE TO MAKE SURE I FOUND THE MAXIMUM. AND THE SECOND DERIVATIVE IS NEGATIVE SO THE FUNCTION DOES LOOK LIKE THAT, SO IT'S A RELATIVE MAX. DOUBLE CHECKING IS ALWAYS A GOOD IDEA. SO I SHOULD ALSO TELL YOU WHAT IS THE VOLUME HERE. SO THE VOLUME IS GOING TO BE 36 TIMES 18 TIMES 18 AND THAT IS GOING TO BE 1164-INCH CUBED. DO YOU REMEMBER WHAT IT WAS NOT THE CYLINDER? THIS WAS THE MAXIMUM VOLUME FOR THIS? DO YOU REMEMBER WHAT THE MAXIMUM VOLUME WAS FOR THIS? 14,000 SOMETHING OR OTHER AM IT WAS BIGGER. SO LET MET JUST APPEAL TO YOUR INTUITION. AND THIS IS THE SAME INTUITION YOU DID BEFORE. SO IF I GIVE YOU A FIXED PERIMETER, A FIX AMOUNT OF STRING AND YOU CAN USE IT TO MAKE ANY SHAME YOU WANT. MAKE A CIRCLE OR SQUARE. SO IT'S A FIX PERIMETER AND I WANT IT PICK THE SHAPE THAT HAS THE BIGGEST AREA. WHAT SHAPE DO YOU THINK HAS THE BIGGEST AREA? A CIRCLE AM YOU GET A CIRCLE YOU CAN YOU DON'T GET A SQUARE WHEN YOU BLOW UP A BALLOON. THAT IS ALWAYS GOING TO GIVE YOU THE BIGGEST ANSWER. I'M JUST A POLING TO YOUR INTUITION THAT YOU SHOULD NOT BE SPRIESED AT HAVING A CIRCLE THERE AND HAVING A BIGGER VOLUME THAN A SQUARE. LET'S DO A DIFFERENT, LET'S BACK UP A SECOND HERE. AND DO A SLIGHTLY EASIER PROBLEM. THAT WAS A HARD ONE. HERE'S WE HAVE'S CHOICE TO MAKE, DEPENDING ON WHICH AVAILABLE. HERE YOU BETTER MAKE THE RIGHT 17 CHOICE OR YOU'RE GOING TO GET IN TROUBLE. SO I WANT TO BUILD A SHELTER WHICH IS JUST LIKE A TENT. AND HERE'S THE SHAPE IS GOING TO BE A BOX LIKE IT WAS BEFORE. I'M GOING IT TELL YOU HAYED OF TIME THE ENDS ARE GOING TO BE SQUARE. AND THE LENGTH IS GOING TO BE Y, AND THIS IS A SHELTER, SO IT'S GOING TO BE SITTING ON THE GROUND SO THERE'S GOING TO BE NO BOTTOM. IT'S JUST OPEN. THERE'S NO FRONT. ONLY GOING TO BE TWO SIDES. A TOP AND BACK WALL. AND WHAT I WANT TO DO IS I WANT TO MAXIMUM MAXIMIZE THE VOLUME OF MY SHELTER. AND SO WHAT'S THE VOLUME GOING TO BE? I'LL WRITE THAT DOWN IN A SECOND. BUT I HAVE A CONSTRAINT. I ONLY HAVE A FIX AMOUNT OF CLOTH OR WHATEVER TO BUILT IT WITH. I HAVE A FIX LIMITED AMOUNT. SO THE CONSTRAINT IS THE AREA OF THE SURFACE. I ONLY HAVE A FIX AMOUNT OF MATERIAL THAT I CAN USE TO BUILD ALL THE SURFACES. SO LET'S, AND THE ANSWER IS THE FIX AREA. SO SURFACE IS GOING TO BE 96 SQUARE FEET. SO I WANT IT BUILD A BIG A SHELTER AS I CAN. SO LET'S WRITE DOWN FORMULAS. SO WHAT IS THE VOLUME OF THIS RECTANGULAR BOX? THIS SIDE IS Y-. THE OTHER TWO SIDES ARE X. Y-X-SQUARED. THAT'S EASY. NOW THE OTHER CONSTRAINT, I'M GOING IT MAKE IT EQUAL, USE ALL THE CLOTH DO MAKE IT AS BIG AS POSSIBLE. SO WHAT IS THE SURFACE AREA? IT ONLY HAS FOUR SIDES AM THE BOTTOM IS OPEN AND THE FRONT. SO WHAT IS MY CONSTRAINT? HOW ABOUT, SO WHAT ARE THE AREA OF THE TWO ENDS? (ON BOARD). WHAT ABOUT THE TOP AND BACKBOARD BOARD. SO THIS PROBLEM LOOKS SIMILAR TO THE ONE I HAD BEFORE. TWO VARIABLES. Y-AND X. I HAVE VOLUME AND AREA SO NOW 18 I HAVE TO DECIDE. DO I SOLVE FOR X-OR DO I SOLVE CONSTRAINT FOR Y. (INAUDIBLE). SOLVING THIS FOR Y-IS EASY, BECAUSE IT'S LINEAR. I COULD SOLVE FOR X-I'D HAVE O TO USE THE QUADRATIC

9 FORMULA AND IT WOULD BE UGLY. SO MUCH EASIER TO SOLVE FOR Y. SO HERE YOU HAVE TO USE JUDGMENT. AND PICK THE ONE THAT MAKE ALGEBRA EASIER. YOU COULD DO IT EITHER WAY BUT PICK THE ONE THAT'S EASIER. PICK ONE AND SOLVE IT. SO I'M GOING TO SOLVE THAT FOR Y. SO GOING TO TAKE 96 MINUS TWO X-SQUARED EQUALS TWO X-Y. MOVE THAT OVER TO THE SIDE AND DIVIDES BOTH SIDES BY TWO X-IN ORDER TO SOLVE FOR Y. SO HERE I GET Y-. AND OVER HERE I GETS 48 DIVIDED BY X-MINUS X. (ON BOARD). SO THAT'S A PRETTY SIMPLE THING. LET'S JUST GO AHEAD AND DO THIS NOW. SO THE VOLUME IS GOING TO BE X-SQUARED TIMES Y-AND THAT'S X-SQUARED TIMES WHERE YOU WROTE DOWN, 48 DIVIDED BY X-MINUS X. AND THAT'S 48 X-MINUS X-CUBED. NICE SIMPLE POLYNOMIAL. FIND THIS CRITICAL POINTS. TAKE THE DERIVATIVE. AND AND SET IT EQUAL TO ZERO. SOLVE. SO THAT'S 48 MINUS THREE X-SQUARED. SO THREE X-SQUARED EQUALS 48 DIVIDED BOTH SIDES BY THREE. AND I GET X-SQUARED EQUAL 16 OR X-EQUALS FOUR OR MINUS FOUR BUT WE KNOW THAT'S NOT A REALISTIC ANSWER. IT'S A LENGTH, IT CAN'T BE A NEGATIVE. THE ONLY SQUARE ROOT THAT MATTER IS A POSITIVE ONE AM I BETTER CHECK FOR THE DERIVATIVE, SECOND DERIVATIVE. MINUS SIX X-. THAT IS NEGATIVE, SO IT'S CONCAVE DOWN SO YES IT IS A MAXIMUM. STUDENT: SIDE OF SQUARES. PROFESSOR: HERE THE STATEMENT OF PROBLEM IS I TOLD YOU IT HAS 19 SQUARE AHEAD OF TIME. I'LL GIVE A HANDWRITTEN ABOUT THE GENERAL CASE. IN THIS CASE, THE PROBLEM WAS STATED FOR SURE YOU START WITH A SQUARE. SO WE GET, SO WE'RE SURE WE FOUND THE MAXIMUM. THE MAXIMUM IS AT POINT X-IS FOUR. SO THAT MEANS THAT Y-IS EQUAL TO 48 DIVIDED BY FOUR MINUS FOUR, IS EIGHT. (ON BOARD). AND SO THE FINAL VOLUME IS X-SQUARED TIMES Y-FOUR SQUARED TIME EIGHT I THINK THAT'S 128 CUBIC FEET WHEN ALL IS SAID AND DOWN. STUDENT: WHAT IS THAT X SQUARED EQUALS THREE X-SQUARED. PROFESSOR: I'M SOLVING THIS THING EQUAL TO ZERO. I SAID 48 EQUAL TO THREE X-SQUARED. AND THEN I DIVIDEED BOTH SIDES BY THREE AND SQUARE ROOT. SOMEBODY ASKED COULD I DO THE MORE GENERAL PROBLEM WHERE IT'S, WHERE I DON'T TELL YOU IT'S A SQUARE, SO IT'S X-BY Y-BY Z. I'M IN THE GOING TO DO THAT PROBLEM BUT I KNOW TO SAY A LITTLE BIT ABOUT IT. THIS COULD BE Y-TIMES X-TIMES Z. SO I WOULD HAVE, SO THIS IS WHAT I WANT TO MAXIMIZE AND HERE WOULD BE THE CONSTRAINT WOULD BE AREA WHICH IS STILL GOING TO BE 96. AND THAT WOULD BE THE TWO ENDS. X-TIMES Z-AND THEN THE TOP WOULD BE Y-TIMES Z. X-TIMES Z-AND THE BACK WOULD BE Y-TIME Z. SO THAT'S, SO WE USE THE SAME TRICK AS BEFORE. GET THREE VARIABLES. I CHOOSE ONE TO HOLD CONSTANT AND SOLVE FOR THE OTHER TWO. AND WE CAN -- STUDENT: THE TOP IS X-Y-NOT X-Z- PROFESSOR: THANK YOU. X-Y. YEAH. OKAY. X-TIMES Y-AND Y-TIMES Z-. THANK YOU. MY ART IS NOT GREAT THERE. SO THE ONLY TROUBLE HERE IS THAT THE REASON I'M NOT GOING TO DO IT IS WE NEED SOME 20 NEW DIFFERENTIATION RULES. WE'RE GOING TO GET SOME FUNCTIONS THAT ARE A LITTLE BIT HARDER. BUT WE'LL SEE THOSE RULES IN THE NEXT CHAPTER. SO REST ASSURED BY THE ENDS OF COURSE YOU WOULD BE ABLE TO SOLVE THIS PROBLEM EVEN IF YOU CAN'T TODAY. SLIGHTLY MORE COMPLICATED FUNCTION BUT THE SAME APPROACH. SO ANY QUESTIONS? I'LL SAVE THIS ONE FOR FUTURE. I'M SURE YOU'LL BE PLEASED.

10 SO MAYBE YOU'RE ALL A LITTLE TIRED OF MAXIMIZING BOXES, SO LET ME CHOOSE ANOTHER EXAMPLE FROM BUSINESS. AND TRY TO SOLVE AN OPTIMIZATION PROBLEM THERE AND THEN I'LL CHOOSE ONE FROM MEDICINE. SO THESE ARE GOING TO BE ELEMENTARY EXAMPLES BUT I SURE YOU THE REAL ONES, SOME OF THE WHICH I KNOW MANY PEOPLE AROUND HERE WORK ON ARE MUCH MORE COMPLICATED AND THEY STILL USE CALCULUS. SO THE FIRST ONE IS CALLED INVENTORY CONTROL. YOU RUN A BUSINESS. YOU HAVE TO ORDER A PRODUCT OF YOU HAVE TO KEEP SUPPLIES AND THE QUESTION IS HOW FREQUENTLY SHOULD YOU ORDER AND REORDER SOME SUPPLY THAT YOU NEED FOR YOUR BUSINESS. SO THAT'S THE HIGH LEVEL QUESTION. LET ME GIVE YOU SOME MORE DETAIL SO IT MAKE SENSE SO WE CAN TURN IT INTO A PROBLEM. SO OF COURSE IT DEPENDS ON HOW FAST DO YOU USE UP THE SUPPLIES. AND TO MAKE THIS PROBLEM DOABLE WE'RE GOING TO ASSUME YOU USE THEM UP AT DISIENT RATE EVERYDAY YOU USE UP OR SELL SAME NUMBER OF SUPPLIES, YOU BUY AND USE EVERYDAY. WHATEVER IT IS. THE OTHER THING THAT YOU WANT TO DO, I HAVEN'T TOLD YOU WHAT THE OBJECTIVE FUNCTION IS, HOW FREQUENTLY SHOULD YOU REORDER SOME SUPPLY YOU NEED TO 21 MINIMIZE THE COST. NOW I'M GOING IT TELL YOU WHY IT COST MONEY TO DO THIS. OF COURSE YOU HAVE TO PAY FOR IT. AND SUPPLIES THEMSELVES BUT THERE'S HAVE A COSTS TOO. THE FIRST ONE IS CALLED A CARRYING COSTS. THAT MEANS IF YOU ORDER SUPPLIES YOU HAVE TO PAY TO STORE THEM AND CARE FOR THEM UNTIL YOU USE THEM UP. SO THAT'S THE COST OF STORAGE, INSURANCE, STUFF LIKE THAT. DEPENDS ON HOW MUCH YOU HAVE STORED. AND I HAVE TO TELL YOU HOW MUCH THAT COST IS PROPORTIONAL TO HOW MUCH, HOW BIG YOUR INVENTORY IS. PROPORTIONAL TO YOUR -- IF YOU KNOW IT MINIMIZE THAT IT TELL YOU YOU DON'T WANT IT KEEP VERY MUCH ON HANDS. BUT THERE'S ANOTHER COST, WHICH IS THE ORDERING COST. AND THIS IS A, WHENEVER YOU ORDER SOMETHING IT HAS TO GO THROUGH THE MAIL, A TRUCK HAS TO SHOW UP. FIX COST TO ORDER TO GET THIS STUFF DELIVERED. SO THIS IS A DIFFERENT KIND OF COST. SO LET'S THINK ABOUT WHAT THAT MEANS. IF ALL YOU CARED ABOUT WAS THE CARRYING COST, SO IF YOU WANT TO MINIMIZE OF THE CARRYING COST ALONE YOU DIDN'T CARRY ABOUT ORDERING, YOU'D ORDER VERY VERY TINY AMOUNTS VERY FREQUENTLY. SO EVERYDAY YOU ORDER EXACTLY WHAT YOU NEEDED THAT DAY. IT WOULD ARRIVE AND YOU HAVE A SMALL INTENSETORY AS POSSIBLE BECAUSE YOU ONLY HAVE ONE DAY A SUPPLY. THAT'S WHAT YOU WOULD DO. BUT IF YOU ORDERED ONCE A DAY YOU HAVE A HIGH ORDERING COST. SO IN ORDER TO MINIMIZE ORDERING COST, ALONG, IF THAT'S ALL YOU CARED ABOUT, YOU'D DO ONE BIG ORDER PER YEAR. SO IN OTHER WORDS USE MAXIMIZE YOUR INVENTORY, JUST ORDER EVERYTHING AT ONCE? 22 A IF YOU DID AT THAT YOU'D HAVE VERY BIG CARRYING COST. IF YOU DID IT YOU'D HAVE BIG ORDERING COST. SO SOMEWHERE IN THE MIDDLE, THERE'S -- SO YOU HAVE TO BALANCE THESE TWO. SO THE OPTIMUM COST WHERE IN THE MEDDLE. SOMEWHERE IN THE MIDDLE IS THE BEST THING AND YOU WANT IT FIND OUT WHAT THAT IS. THAT'S THE PROBLEM SETUP IN WORDS. NOW I'M GOING TO -- THAT'S A WORD PROBLEM. NOW I HAVE TO CHANGE IT INTO A MATH PROBLEM. AND THERE'S SOME BUSINESS JARGON THAT GOES ALONG WITH THIS. THE OPTIMUM ORDER SIZE WHICH WE'RE GOING TO FIGURE OUT IN A MOMENT IS CALLED THE EOQ. JUST CALL THE ECONOMIC ORDER QUANTITY, FANCY WORD. JUST THE ANSWER TO THIS OPTIMIZATION PROBLEM. WE'RE

11 GOING TO FIGURE THAT OUT. SO LET ME DO A CONCRETE EXAMPLE. WHICH IS ORDERING SOME SPECIFIC STUFF. SOME JUICE. AND THEN I'LL DO TOTALLY GENERAL. VARIABLE THAT YOU CAN PLUG IN, DEPENDING ON WHAT YOUR BUSINESS WAS. SO JUST THINK ABOUT ONE YEAR'S WORTH OF BUSINESS. SOME KIND OF FROZEN JUICE. THAT YOU NEED TO HAVE ON HAND IN ORDER TO SELL IN YOUR STORE. AND TURNS OUT WHAT YOU INSIDE IS 1200 CASES. THAT'S ABOUT HOW MUCH YOU SELL IN A YEAR. AND IT'S, IT'S GOING TO BE SOLD AT A CONTINUOUS RATE. EVERYDAY YOU SELL 1200 DIVIDED BY 365 CASES OF JUICE ROUGHLY. SO HERE'S GOING TO BE THE CARRYING COST. IT'S GOING TO BE $8 PER CASE. PER YEAR. IF YOU HAVE ONE CASE SITTING IN STORAGE ALL YEAR IT'S GOING TO COST YOU EIGHT BUCKS. AND THE ORDER COST IS GOING TO BE $75 PER ORDER. NO MATTER HOW MUCH YOU ORDER. THAT'S ALL THE DATA THAT WE NEED. 23 NOW WE'RE GOING TO TRY TO FIGURE OUT HOW MANY ORDERS WE SHOULD MAKE. SO HOW MANY ORDERS PER YEAR MINIMIZE THE TOTAL COST. SO LET ME DRAW SOME PICTURE AGAIN. WE ALL START THESE BY DRAWING PICTURE TO ILLUSTRATE WHAT'S GOING ON. SO LET'S SUPPOSE I DO ONE ORDER PER YEAR. LET ME DRAW THE PICTURE OF WHAT HAPPENS. SO I'M, THIS IS GOING TO BE TIME ALONG THIS AXIS. AND HERE'S GOING TO BE ONE YEAR. SO THIS IS JANUARY 1ST, THERE IS DECEMBER 31ST ALONG THAT AXIS. AND ALONG THE VERTICAL AXIS IS GOING TO BE WHAT YOUR INVENTORY IS, HOW MUCH YOU ACTUALLY HAVE. ON JANUARY 1ST IF YOU ORDER ONE PER YEAR YOU'RE GOING TO ORDER 1200 CASES. THEY'RE ALL GOING TO SHOW UP AND YOUR INVENTORY IS GOING TO BE (INAUDIBLE). SELL EACH EQUAL, INVENTORY IS GOING TO BE STRAIGHT LINE. IT'S GOING TO HIT ZERO THERE. THAT'S GOING TO BE IN\CONVENIENT\VEINTORY AM SELL EQUAL AMOUNT EVERYONE DAY. GOES DOWN IN A NICE STRAIGHT LINE. IF I HAD THAT PICTURE WHAT IS GOING TO BE THE TOTAL COST. I WANT TO WRITE THAT DOWN. AND COMPARE TO OTHERS. SO THERE'S GOING TO BE THE TWO PIECES. GOING TO BE THE CARRYING COST. AND THE ORDERING COST. AND TO MAKE IT EASY, ON THE FIRST DAY YOU HAVE 1200 CASES, NEXT DAY, AND SO YOU'VE PAID EIGHT BUCKS PER CASE. ON THAT. SO TO MAKE IT EASY WHAT I'M GOING TO DO IS SAY WHAT IS THE TOTAL INVENTORY COST WILL DEPENDS ON THE AVERAGE INVENTORY FOR THE YEAR. IF YOU START WITH 1200, AND EVERYDAY YOU SELL AN EQUAL AMOUNT YOU END UP WITH ZERO AN AVERAGE WHAT'S YOUR INVENTORY? HALF. AVERAGE INCONVENIENT STOAR IS 600 CASES. SO THE INVENTORY COST IS GOING 24 TO BE 600 CASES, THAT'S THE AVERAGE THAT YOU HAVE IN. AND IT'S $8 A CASE. SO THAT'S GOING TO BE IN\CONVENIENT\VEINTORY COST. AND WHAT IS THE ORDERING COST? YOU ORDER ONCE, THERE'S ONE ORDER, THAT COSTS 75 BUCKS. AND SO THE TOTAL COST, TOTAL CARRYING COST I'M GOING TO ADD THOSE TWO UP AND I'LL GET $4,075. THAT'S ONE SCENARIO THAT WE CAN DO. I HAVEN'T SEEN NUCLEUS YET. DO THIS BY HANDS A COUPLE EXAMPLES AND THEN TURN INTO CALCULUS PROBLEMS. STUDENT: TAKE THE AVERAGE. PROFESSOR: SO THE WAY YOU WOULD FIGURE IT OUT IS YOU SAY, IT'S EIGHT BUCK PER CASE PER YEAR, SO IT WOULD BE, SO EIGHT DIVIDED BY 365 PER CASE TODAY. AND THEN YOU SAY, I HAVE THIS MANY CASES ON DAY TWO AND THIS MANY ON DAY TWO, IT'S EASIER U-YOU GET THE SAME ANSWER BY SAYING WHAT'S THE AVERAGE AND I'LL JUST PAY EIGHT BUCKS TIMES THE AVERAGE. YOU GET THE SAME ANSWER. I'M SIMPLIFYING.

12 STUDENT: ONE. PROFESSOR: ONE ORDER PER YEAR. SO THIS IS ONE ORDER PER YEAR ON JANUARY 1ST YOU ORDERS ENTIRE YEAR'S SUPPLY WHICH IS 1200 THEN IT GO DOWN CONTINUOUSLY UNTIL DECEMBER 31ST YOU SELL YOUR LAST CASE AND THEN YOU ORDER AGAIN. SO SORT OF THE, HAD A MINIMIZESING ORDERING COST. YOU HAVE TO HAVE AT LEAST ONE ORDER BY MAXIMIZES THAT. THE NEXT OBVIOUS POSSIBILITY IS I CAN DO TWO ORDERS PER YEAR. SO IF, HOW BIG WOULD EACH ORDERS BE THEN IF I NEED 1200 CASES AND I DO TWO ORDERS? 25 STUDENT: I'M CONFUSEDDED AS TO WHY YOU CHOOSE THE AVERAGE INVENTORY. PROFESSOR: FOR EVERY, I DIDN'T WANT TO DO ALL THE ALGEBRA TO EXPLAIN IT BUT YOU GET THE SAME ANSWER IF YOU DO COST PER DAY AS IF YOU SAY WHAT IS THE AVERAGE OVER THE WHOLE YEAR WHICH IS 600. AND PAY EIGHT. AND MULTIPLY THAT BY YOUR AFFIRM, BY THE COST PER YEAR. SO MAYBE FOR THE SAKE OF ARGUMENT IT'S GOING TO BE THE AVERAGE INVENTORY TIMES EIGHT. SO ALL THESE OTHER SCENARIOS. LET ME DO THE NEXT ONE AND YOU'LL SEE -- AGAIN DRAW THE SAME PICTURE, UP TO A YEAR. INVENTORY AM BUT NOW DO TWO ORDERS AM EACH ORDER IS GOING TO BE FOR 600. GOING IT RUN OUT AFTER SIX MONTHS. SO START THERE. BUT I'M SO I HAVE TO MAKE ANOTHER ORDER AM GET ANOTHER 600 CASES. AND I'LL RUN OUT AFTER ONE YEAR. SO THAT'S WHAT INVENTORY LOOK LIKE. TWO ORDERS THE SIZE OF 600. WHAT IS THE AVERAGE NOW THAT I HAVE IN INVENTORY IF THAT IS 600? THE AVERAGE EQUALS 300. SO THE TOTAL COST IS GOING TO BE 300, THE AVERAGE INVENTORY TIMES EIGHT BUCKS FOR A CASE, PLUS HOW MANY ORDERS DID I JUST MAKE. PROFESSOR: TWO TIME 75. SO THAT'S THAT'S HOW MUCH IT IS. NEXT POSSIBILITY. LET ME DO, MAKE SURE IT'S COMPLETELY FIRM IN YOUR MIND. FOUR ORDER PER YEAR AM SAME OLD PICTURE. TIME. VERSUS INVENTORY. AND IF I DO FOUR ORDERS PER YEAR, HOW BIG ARE THE ORDERS? 300. HOW MANY TIME AM I GOING TO DO IT? FOUR TIMES. SO THIS IS WHAT THE INVENTORY IS GOING TO LOOK LIKE. WHAT IS THE AVERAGE INVENTORY GOING TO BE? 150. SO THE COST IS 26 GOING TO BE 150 TIMES EIGHT PLUS FOUR TIME 75. AND IF YOU GET BETTER, OKAY. SO IT SEEM LIKE IT'S MORE ORDERS IS BETTER. SO LET'S JUST IMAGINE NOW THAT WE GET ONE ORDERS AM I'M NOT GOING TO DRAW THE PICTURE. DON'T WORRY. WHAT'S THE COST GOING TO BE? WITHOUT DRAWING THE PICTURE WHAT IS THE COST GOING TO BE? NOW I'M GOING TO DO 100 ORDERS PER YEAR AM SO THERE'S GOING, I NEED IT MULTIPLY THE AVERAGE INVENTORY TIMES EIGHT. SO WHAT IS THE SIESES OF EACH ORDERS GOING TO BE? I HAVE 1200, I NEED 1200 ALL YEAR. SO THE TOTAL, SIZE OF EACH ORDER IS 12, 1200 DIVIDED BY 100. SO THE AGE IS TWO SO THAT'S SIX. AND THEN THERE ARE GOING TO BE 100 ORDERS AM EACH ONE COSTING 75 BUCK. IF I, OPTIMUM IS SOMEWHERE IN BETWEEN SO I WANT IT CHANGE NOW AND DO THE CALCULUS PROBLEM AND FIGURE OUT WHAT THE OPTIMUM. COST WENT DOWN FOR AWHILE AS I INCREASED BUT I CAN'T AFFORD. STUDENT: THAT'S ASSUMING THAT YOU'RE RUNNING OUT OF INVENTORY IN A CERTAIN AMOUNT OF TIME. PROFESSOR: I'M ASSUMING, NUMBER EVERYDAY. I SELL AN EQUAL NUMBER OF CASES EVERYDAY. SO MAKE THIS PROBLEM A LITTLE BIT EASIER. IF YOU HAD SEASONAL SALES IT WOULD STILL BE A CALCULUS

13 PROBLEM BUT IT WOULD BE MORE COMPLICATED TO WRITE DOWN. SO LET ME WRITE IT DOWN AS A CALCULUS PROBLEM. HERE'S THE GENERAL CASE. NOW I'M GOING IT GIVE NAMES TO ALL THE QUANTITY. CALL R-NUMBER OF ORDER PER YEAR. I'LL CALL X-THE ORDERS SIZE. SO WHAT IS MY CONSTRAINT, IF I SELL 1200 CASES A YEAR, HOW MUCH DO I HAVE TO ORDER? TOTAL NUMBER OF CASES THAT 27 YOU NEED HAS GOT TO BE SOMETHING TIMES SOMETHING. DO R-ORDER PER YEAR AND EACH ONE HAS X-CASES IN THE ORDERS HOW MANY HAVE I ORDERED? R-TIMES X. R-ORDER PER YEAR TIME X-CASES PER ORDER. THAT'S HOW MANY CASES I ORDERED PER YEAR AND MY CONSTRAINT IS IT HAS TO BE THAT'S WHAT I NEED. I HAVE TO ORDER 1200 CASES. SO NOW LET ME WRITE DOWN COST. THE THING I WANT TO MINIMIZE. SO IT'S GOING TO BE, LET'S BE ORDER COST. I HAVE R ORDER PER YEAR, EACH ORDER COST 75 BUCKS. SO (ON BOARD). AND WHAT ABOUT THE OTHER COST? SO I HAVE TO FIGURE OUT THE AVERAGE INVENTORY AM SO THE MAXIMUM I HAVE IN STOCK IS X. SO WHAT IS, AND THEN IT GOES DOWN CONTINUOUSLY. SO THE AVERAGE IS? THE AVERAGE IS X-OVER TWO AND MULTIPLY THAT BY EIGHT. SO FOUR X, YES. SO THIS IS THE AVERAGE INVENTORY. SO THAT COMES OUT TO 75 R-PLUS FOUR X. (ON BOARD) SO NOW IT'S JUST THE SAME PROBLEM THAT I HAD BEFORE. SO I WANT IT MINIMIZE THIS COST. SO LET'S GO AHEAD AND DO THAT. LET ME SOLVE THE CONSTRAINT FOR ANY, EITHER ONE VARIABLE WILL DO. SO X-IS 1200 DIVIDED BY R. SO SOLVE THE CONSTRAINT EQUATION. SUBSTITUTE THAT IN HERE. AND I'LL GET THE COST IS GOING TO BE 75 R-PLUS FOUR TIMES 1200 OVER X-AND THAT'S 75, SORRY OVER R. SEVENTY-FIVE R-PLUS 4800 OVER R. SO NOW I HAVE COST IN TERMS OF ONE VARIABLE. I CAN DO NOW CALCULUS. TAKE THE DERIVATIVE. SO DO C-PRIME DIFFERENTIATE WITH RESPECT TO R. IT'S 75 TIMES, WHAT IS NEXT? MINUS (ON BOARD). DIVIDED BY WHAT? R-SQUARED. SO I WILL SOLVE, SO I WANT TO FIND THE CRITICAL POINT. (ON BOARD) THAT MEANS R-IS THE SQUARE ROOT OF 28 64, WHICH IS NOT GOING TO BE NEGATIVE EIGHT. IT'S GOING TO BE POSITIVE EIGHT. SO PICK THE SENSIBLE SOLUTION. SO LET'S JUST DOUBLE CHECK IF IT'S A MAXIMUM OR MINIMUM. SO LET MET DIFFERENTIATE THIS WITH RESPECT TO R-ONE MORE TIME. C-PRIME. I WANT C-DOUBLE PRIME OVER R-CUBED. HERE'S AND THIS IS POSITIVE OR NEGATIVE? IT'S ALWAYS POSITIVE. IT'S ALWAYS CONCAVE UP. SO IT'S A (INAUDIBLE). SO THAT MEANS I FOUND THE MINIMUM. IT'S EIGHT ORDER PER YEAR. OKAY. SO NEXT TIME I WILL REVISIT IT QUESTION. AND DO IT IN GENERALITY FOR ANY ORDER SIZE AND ANY COST.

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