- 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.