Ex: Optimization – Minimize the Cost to Make a Can with a Fixed Volume
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– A CYLINDER-SHAPED CAN
NEEDS TO BE CONSTRUCTED TO HOLD 300 CUBIC CENTIMETERS
OF SOUP. THE MATERIAL
FOR THE SIZE OF THE CAN COSTS 4 CENTS
PER SQUARE CENTIMETER. THE MATERIAL FOR THE TOP
AND BOTTOM OF THE CAN NEEDS TO BE THICKER AND COST
5 CENTS PER SQUARE CENTIMETER. WE WANT TO FIND THE DIMENSIONS
FOR THE CAN THAT WILL MINIMIZE
THE PRODUCTION COSTS. SO THE FIRST THING WE SHOULD
RECOGNIZE ABOUT THIS PROBLEM IS THAT THE VOLUME OF THE CAN
OR THE RIGHT CIRCULAR CYLINDER MUST BE 300 CUBIC CENTIMETERS AND SINCE THE VOLUME
OF A CYLINDER IS EQUAL TO PI R SQUARED H, FOR THIS PROBLEM PI R SQUARED H
MUST EQUAL 300 CUBIC CENTIMETERS AND THIS IS CALLED
THE CONSTRAINT. NEXT, THE COST FUNCTION INVOLVES
THE SURFACE AREA OF THE CYLINDER AS WELL AS DIFFERENT COSTS FOR THE TOP, BOTTOM,
AND SIDE OF THE CAN. SO LETS FIRST TALK ABOUT THE
SURFACE AREA OF OUR CYLINDER. IF YOU CAN IMAGINE CUTTING OUT
THE TOP AND BOTTOM OF THE CAN AND THEN SLICING THE SIDE
AND UNROLLING THE SIDE HOPEFULLY YOU CAN VISUALIZE THAT IT WOULD LOOK SOMETHING
LIKE THIS WHERE THIS WOULD BE THE TOP
AND THE BOTTOM AND THIS WOULD BE THE SIDE
OF THE CAN. NOTICE HOW THE RADIUS FOR THE
TWO CIRCLES AND THE CYLINDER WOULD BE THE SAME. THE HEIGHT OF THE CAN WOULD GIVE
US THIS LENGTH OF THE RECTANGLE AND THEN THIS LENGTH HERE
IS A LITTLE TRICKY. THIS ACTUALLY COMES FROM THE
CIRCUMFERENCE OF THE CIRCLE HERE, AND THE FORMULA FOR THE
CIRCUMFERENCE OF A CIRCLE IS EQUAL TO 2 PI R AND AGAIN, REMEMBER THE COST FOR
THE TOP AND BOTTOM IS 5 CENTS
PER SQUARE CENTIMETER, AND THE COST OF THE SIDE IS
4 CENTS PER SQUARE CENTIMETER. SO WITH THIS INFORMATION
WE CAN WRITE THE COST EQUATION. THE COST IS GOING TO BE EQUAL
TO 5 CENTS x THE AREA OF THE TWO CIRCLES. WELL THE AREA OF A CIRCLE
IS EQUAL TO PI R SQUARED. SO WE’D HAVE x 2 PI R SQUARED
+ 4 CENTS x THE AREA OF THE RECTANGLE
WHICH WOULD BE 2 PI R H. SO LET’S GO AHEAD
AND TAKE THIS COST EQUATION BACK TO THE PREVIOUS SLIDE. REMEMBER OUR GOAL HERE
IS TO MINIMIZE THE COST, BUT WE WANT THE COST FUNCTION
IN TERMS OF ONE VARIABLE AND RIGHT NOW IT’S IN TERMS
OF TWO VARIABLES BOTH R AND H. SO WHAT WE’LL DO NOW
IS USE THE CONSTRAINT TO PERFORM A SUBSTITUTION. SO THAT WE CAN OBTAIN THE COST
FUNCTION IN TERMS OF JUST R. TO DO THIS WE’RE GOING TO SOLVE
THE CONSTRAINT FOR H. SO WE’RE GOING TO DIVIDE
BOTH SIDES OF THIS EQUATION BY PI R SQUARED. SO H=300 DIVIDED BY
PI R SQUARED. NOW WE CAN SUBSTITUTE
THIS QUOTIENT HERE FOR H IN OUR COST EQUATION. SO WE WOULD HAVE C=0.05 x 2
IS 0.1. SO WE HAVE 0.1 PI R SQUARED
+ 0.04 x 2 WOULD BE 0.08 PI x R x H BUT AGAIN H IS EQUAL TO 300
DIVIDED BY PI R SQUARED AND NOW LETS GO AND SIMPLIFY
THIS PRODUCT HERE. PI/PI WOULD SIMPLIFY TO 1. NOTICE ONE FACTOR OF R
WOULD SIMPLIFY AS WELL. THIS WOULD SIMPLIFY TO 1. THIS WOULD SIMPLIFY
R TO THE FIRST. SO NOW WE HAVE THE COST FUNCTION
IN TERMS OF R. WE HAVE C=0.1 PI R SQUARED
+ 0.08 x 300 IS 24. SO WE’D HAVE 24 DIVIDED BY R WHICH WE CAN ALSO WRITE AS
24 R TO THE POWER OF -1. REMEMBER IF WE MOVE R
UP INTO THE DENOMINATOR OR ACROSS THE FRACTION BAR IT’S GOING TO CHANGE THE SIGN
OF THE EXPONENT. SO NOW THAT WE HAVE OUR COST
FUNCTION IN TERMS OF R TO MINIMIZE C WE’LL FIND THE
CRITICAL NUMBERS BY DETERMINING WHERE THE FIRST
DERIVATIVE IS EQUAL TO ZERO OR UNDEFINED AND THEN FROM THERE WE’LL
DETERMINE WHICH VALUE OF R MINIMIZES THE COST. SO C PRIME WHICH WILL BE
THE DERIVATIVE WITH RESPECT TO R WOULD BE 0.2 PI x R TO THE FIRST
OR JUST R AND HERE WE’RE GOING TO MULTIPLY
BY -1 SO WE’LL HAVE – 24 R
TO THE POWER OF -1 – 1 OR -2. LET’S GO AHEAD AND TAKE THIS
DERIVATIVE FUNCTION TO THE NEXT SLIDE TO DETERMINE WHERE IT’S
UNDEFINED OR EQUAL TO ZERO. LET’S GO AHEAD AND REWRITE THIS
AS A FRACTION. WE CAN WRITE THIS AS 0.2 PI R
– 24 DIVIDED BY R SQUARED EQUALS ZERO. NOW WE CAN SEE IT’S GOING TO BE
UNDEFINED WHEN R=0 BUT WE KNOW THE RADIUS
CAN’T EQUAL 0 THEREFORE WE’LL JUST GO AHEAD
AND SOLVE THIS FOR R BY MULTIPLYING BOTH SIDES
OF THE EQUATION BY R SQUARED. SO ON THE LEFT SIDE WE’RE GOING
TO HAVE 0.2 PI R CUBED. NOTICE HOW THE R SQUAREDS
WILL SIMPLIFY OUT SO WE’LL HAVE – 24=0. WE’LL GO AHEAD AND SOLVE THIS
FOR R CUBED SO WE’LL ADD 24 TO BOTH SIDES AND WE’LL GO AHEAD AND DIVIDE BY
0.2 PI. SO WE HAVE R CUBED=24
DIVIDED BY 0.2 PI. NOW TO SOLVE FOR R WE’LL TAKE THE CUBE ROOT OF BOTH
SIDES OF THE EQUATION. TO SAVE SOME TIME I’VE ALREADY
DETERMINED THIS VALUE. WE HAVE R IS APPROXIMATELY EQUAL
TO 3.3678. NOW BECAUSE WE ONLY HAVE
ONE CRITICAL NUMBER WE’D PROBABLY WOULD BE OKAY
BY ASSUMING THIS VALUE OF R IS GOING TO
MINIMIZE THE PRODUCTION COST. BUT JUST TO BE SURE WE SHOULD
USE EITHER THE FIRST OR SECOND DERIVATIVE TEST TO MAKE SURE WE DO HAVE
A MINIMUM VALUE AT THIS VALUE OF R. LET’S GO AHEAD AND USE
THE SECOND DERIVATIVE TEST. SO IF HERE’S THE FIRST
DERIVATIVE THEN WE CAN FIND
THE SECOND DERIVATIVE. APPLYING THE POWER RULE HERE
WE WOULD JUST HAVE 0.2 PI. THEN WE’D MULTIPLY BY -2 SO THAT’S GOING TO BE + 48
x R TO THE POWER OF -3 WHICH IS THE SAME AS DIVIDING BY
R TO THE 3rd. NOW NOTICE IF WE SUBSTITUTED
3.3678 FOR R THIS SUM WOULD BE POSITIVE. SO C DOUBLE PRIME OF 3.3678
WOULD BE GREATER THAN ZERO WHICH MEANS THE FUNCTION WOULD
BE CONCAVE UP WHICH IS GOOD NEWS BECAUSE IF THE FUNCTION
IS CONCAVE UP AT THIS CRITICAL NUMBER THAT MEANS WE’D HAVE A MINIMUM
FUNCTION VALUE OR IN THIS CASE
A MINIMUM PRODUCTION COST. WE’RE GOING BACK TO THE PROBLEM
JUST FOR A MOMENT. WE’RE TRYING TO FIND
THE DIMENSIONS THAT WILL MINIMIZE THE COST. SO WE JUST FOUND THE RADIUS
THAT WILL MINIMIZE THE COST BUT WE STILL HAVE TO FIND
THE HEIGHT OF THE CYLINDER THAT WOULD MINIMIZE THE COST AND WE CAN DO THAT
USING THIS EQUATION HERE SINCE WE KNOW H MUST EQUAL 300
DIVIDED BY PI R SQUARED. SO LET’S GO AHEAD AND FINISH
BY DOING THAT SO THAT MEANS H IS EQUAL TO 300
DIVIDED BY PI x 3.3678 TO THE 2nd POWER. AGAIN, TO SAVE SOME TIME I’VE
ALREADY DETERMINED THIS VALUE. H COMES OUT TO APPROXIMATELY
8.4194 CENTIMETERS. SO WE KNOW THE PRODUCTION COSTS
UNDER THESE CONDITIONS IS MINIMIZED IF THE CAN HAS A RADIUS
OF 3.3678 CENTIMETERS AND A HEIGHT OF APPROXIMATELY
8.4194 CENTIMETERS. NOW OF COURSE IF ANY
OF THE CONDITIONS CHANGE, FOR EXAMPLE IF THE VOLUME
CHANGES OR IF THE COSTS FOR THE SIDE
OR THE TOPS OF THE CAN CHANGE THE DIMENSIONS OF THE CAN THAT WOULD MINIMIZE
THE PRODUCTION COSTS WOULD ALSO CHANGE. I HOPE YOU FOUND THIS
EXPLANATION HELPFUL.  

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