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- AN OPEN TOP RECTANGULAR BOX IS BEING CONSTRUCTED
TO HOLD A VOLUME OF 350 CUBIC INCHES.
THE BASE OF THE BOX IS MADE FROM MATERIAL
COSTING 6 CENTS PER SQUARE INCH.
THE FRONT OF THE BOX MUST BE DECORATED
AND WILL COST 12 CENTS PER SQUARE INCH.
THE REMAINDER OF THE SIDES WILL COST 2 CENTS PER SQUARE INCH.
FIND THE DIMENSIONS THAT WILL MINIMIZE
THE COST OF CONSTRUCTING THIS BOX.
LET'S FIRST DIAGRAM THE BOX AS WE SEE HERE
WHERE THE DIMENSIONS ARE X BY Y BY Z
AND BECAUSE THE VOLUME MUST BE 350 CUBIC INCHES
WE HAVE A CONSTRAINT THAT X x Y x Z MUST EQUAL 350.
BUT BEFORE WE TALK ABOUT OUR COST FUNCTION LETS TALK ABOUT
THE SURFACE AREA OF THE BOX.
BECAUSE THE TOP IS OPEN, WE ONLY HAVE 5 FACES.
LET'S FIND THE AREA OF THE 5 FACES
THAT WOULD MAKE UP THE SURFACE AREA.
NOTICE THE AREA OF THE FRONT FACE WOULD BE X x Z
WHICH WOULD ALSO BE THE SAME AS THE AREA IN THE BACK
SO THE SURFACE AREA HAS TWO XZ TERMS.
NOTICE THE RIGHT SIDE OR THE RIGHT FACE
WOULD HAVE AREA Y x Z WHICH WILL BE THE SAME AS THE LEFT.
SO THE SURFACE AREA CONTAINS TWO YZ TERMS
AND THEN FINALLY THE BOTTOM HAS AN AREA OF X x Y
AND BECAUSE THE TOP IS OPEN WE ONLY HAVE ONE XY TERM
IN THE SURFACE AREA AND NOW WE'LL CONVERT THE SURFACE AREA
TO THE COST EQUATION.
BECAUSE THE BOTTOM COST 6 CENTS PER SQUARE INCH
WHERE THE AREA OF THE BOTTOM IS X x Y
NOTICE HOW FOR THE COST FUNCTION
WE MULTIPLY THE XY TERM BY 6 CENTS
AND BECAUSE THE FRONT COSTS 12 CENTS PER SQUARE INCH
WHERE THE AREA OF THE FRONT WOULD BE X x Z
WE'LL MULTIPLY THIS XZ TERM BY 12 CENTS IN THE COST FUNCTION.
THE REMAINING SIDES COST 2 CENTS PER SQUARE INCH
SO THESE THREE AREAS ARE ALL MULTIPLIED BY 0.02 OR 2 CENTS.
COMBINING LIKE TERMS WE HAVE THIS COST FUNCTION HERE.
BUT NOTICE HOW WE HAVE THREE UNKNOWNS IN THIS EQUATION
SO NOW WE'LL USE A CONSTRAINT TO FORM A COST EQUATION
WITH TWO VARIABLES.
IF WE SOLVE OUR CONSTRAINT FOR X BY DIVIDING BOTH SIDES BY YZ
WE CAN MAKE A SUBSTITUTION FOR X INTO OUR COST FUNCTION
WHERE WE CAN SUBSTITUTE THIS FRACTION HERE FOR X
HERE AND HERE.
IF WE DO THIS, WE GET THIS EQUATION HERE
AND IF WE SIMPLIFY NOTICE HOW THE FACTOR OF Z SIMPLIFIES OUT
AND HERE FACTOR OF Y SIMPLIFIES OUT.
SO FOR THIS FIRST TERM IF WE FIND THIS PRODUCT
AND THEN MOVE THE Y UP WE WOULD HAVE 49Y TO THE -1
AND THEN FOR THE LAST TERM IF WE FOUND THIS PRODUCT
AND MOVED THE Z UP WE'D HAVE + 21Z TO THE -1.
SO NOW OUR GOAL IS TO MINIMIZE THIS COST FUNCTION.
SO FOR THE NEXT STEP WE'LL FIND THE CRITICAL POINTS.
CRITICAL POINTS ARE WHERE THE FUNCTION IS GOING TO HAVE
MAX OR MIN FUNCTION VALUES AND THEY OCCUR WHERE THE FIRST ORDER
OF PARTIAL DERIVATIVES ARE BOTH EQUAL TO ZERO
OR WHERE EITHER DOES NOT EXIST.
THEN ONCE WE FIND THE CRITICAL POINTS,
WE'LL DETERMINE WHETHER WE HAVE A MAX OR A MIN VALUE
USING OUR SECOND ORDER OF PARTIAL DERIVATIVES.
SO ON THIS SLIDE WE'RE FINDING
BOTH THE FIRST ORDER AND SECOND ORDER OF PARTIAL DERIVATIVES.
WE HAVE TO BE A LITTLE CAREFUL HERE THOUGH
BECAUSE OUR FUNCTION IS A FUNCTION OF Y AND Z
NOT X AND Y LIKE WE'RE USED TO.
SO FOR THE FIRST PARTIAL WITH RESPECT TO Y
WE WOULD DIFFERENTIATE WITH RESPECT TO Y TREATING Z
AS A CONSTANT WHICH WOULD GIVE US THIS PARTIAL DERIVATIVE HERE.
FOR THE FIRST PARTIAL WITH RESPECT TO Z
WE WOULD DIFFERENTIATE WITH RESPECT TO Z
AND TREAT Y AS A CONSTANT WHICH WOULD GIVE US
THIS FIRST ORDER OF PARTIAL DERIVATIVE.
NOW USING THESE FIRST ORDER OF PARTIAL DERIVATIVES
WE CAN FIND THESE SECOND ORDER OF PARTIAL DERIVATIVES
WHERE TO FIND THE SECOND PARTIALS WITH RESPECT TO Y
WE WOULD DIFFERENTIATE THIS PARTIAL DERIVATIVE
WITH RESPECT TO Y AGAIN GIVING US THIS.
THE SECOND PARTIAL WITH RESPECT TO Z
WE WOULD DIFFERENTIATE THIS PARTIAL DERIVATIVE
WITH RESPECT TO Z AGAIN GIVING US THIS.
NOTICE HOW IT'S GIVEN USING A NEGATIVE EXPONENT
AND IN FRACTION FORM
AND THEN FINALLY FOR THE MIXED PARTIAL
OR THE SECOND ORDER OF PARTIAL WITH RESPECT TO Y
AND THEN Z WE WOULD DIFFERENTIATE THIS PARTIAL
WITH RESPECT TO Z WHICH NOTICE HOW IT WOULD JUST GIVE US 0.04.
SO NOW WE'RE GOING TO SET
THE FIRST ORDER OF PARTIAL DERIVATIVES EQUAL TO ZERO
AND SOLVE AS A SYSTEM OF EQUATIONS.
SO HERE ARE THE FIRST ORDER OF PARTIALS SET EQUAL TO ZERO.
THIS IS A FAIRLY INVOLVED SYSTEM OF EQUATIONS
WHICH WE'LL SOLVE USING SUBSTITUTION.
SO I DECIDED TO SOLVE THE FIRST EQUATION HERE FOR Z.
SO I ADDED THIS TERM TO BOTH SIDES OF THE EQUATION
AND THEN DIVIDED BY 0.04 GIVING US THIS VALUE HERE FOR Z
BUT IF WE FIND THIS QUOTIENT
AND MOVE Y TO THE -2 TO THE DENOMINATOR
WE CAN ALSO WRITE Z AS THIS FRACTION HERE.
NOW THAT WE KNOW Z IS EQUAL TO THIS FRACTION,
WE CAN SUBSTITUTE THIS FOR Z INTO THE SECOND EQUATION HERE.
WHICH IS WHAT WE SEE HERE
BUT NOTICE HOW THIS IS RAISED TO THE EXPONENT OF -2
SO THIS WOULD BE 1,225 TO THE -2 DIVIDED BY Y TO THE -4.
SO WE CAN TAKE THE RECIPROCAL WHICH WOULD GIVE US
Y TO THE 4th DIVIDED BY 1,500,625 AND HERE'S THE 21.
NOW THAT WE HAVE AN EQUATION WITH JUST ONE VARIABLE Y
WE WANT TO SOLVE THIS FOR Y.
SO FOR THE FIRST STEP, THERE IS A COMMON FACTOR OF Y.
SO Y = 0 WOULD SATISFY THIS EQUATION
AND WOULD BE A CRITICAL POINT
BUT WE KNOW WE'RE NOT GOING TO HAVE A DIMENSION OF ZERO
SO WE'LL JUST IGNORE THAT VALUE
AND SET THIS EXPRESSION HERE EQUAL TO ZERO AND SOLVE
WHICH IS WHAT WE SEE HERE.
SO WE'RE GOING TO ISOLATE THE Y CUBED TERM
AND THEN CUBE ROOT BOTH SIDES OF THE EQUATION.
SO IF WE ADD THIS FRACTION TO BOTH SIDES OF THE EQUATION
AND THEN CHANGE THE ORDER OF THE EQUATION
THIS IS WHAT WE WOULD HAVE AND NOW FROM HERE TO ISOLATE Y CUBED
WE HAVE TO MULTIPLY BY THE RECIPROCAL
OF THIS FRACTION HERE.
SO NOTICE HOW THE LEFT SIDE SIMPLIFIES JUST Y CUBED
AND THIS PRODUCT HERE IS APPROXIMATELY THIS VALUE HERE.
SO NOW TO SOLVE FOR Y WE WOULD CUBE ROOT
BOTH SIDES OF THE EQUATION
OR RAISE BOTH SIDES OF THE EQUATION TO THE 1/3 POWER
AND THIS GIVES Y IS APPROXIMATELY 14.1918,
AND NOW TO FIND THE Z COORDINATE OF THE CRITICAL POINT
WE CAN USE THIS EQUATION HERE WHERE Z = 1,225
DIVIDED BY Y SQUARED WHICH GIVES Z IS APPROXIMATELY 6.0822.
WE DON'T NEED IT RIGHT NOW
BUT I WENT AHEAD AND FOUND THE CORRESPONDING X VALUE
AS WELL USING OUR VOLUME FORMULA SOLVE FOR X.
SO X WOULD BE APPROXIMATELY 4.0548.
BECAUSE WE ONLY HAVE ONE CRITICAL POINT
WE CAN PROBABLY ASSUME THIS POINT IS GOING TO MINIMIZE
THE COST FUNCTION BUT TO VERIFY THIS
WE'LL GO AHEAD AND USE THE CRITICAL POINT
AND THE SECOND ORDER OF PARTIAL DERIVATIVES JUST TO MAKE SURE.
MEANING WE'LL USE THIS FORMULA HERE FOR D
AND THE VALUES OF THE SECOND ORDER OF PARTIAL DERIVATIVES
TO DETERMINE WHETHER WE HAVE A RELATIVE MAX OR MIN
AT THIS CRITICAL POINT WHEN Y IS APPROXIMATELY 14.19
AND Z IS APPROXIMATELY 6.08.
HERE ARE THE SECOND ORDER OF PARTIALS THAT WE FOUND EARLIER.
SO WE'LL BE SUBSTITUTING THIS VALUE FOR Y
AND THIS VALUE FOR Z INTO THE SECOND ORDER OF PARTIALS.
WE SHOULD BE A LITTLE CAREFUL THOUGH
BECAUSE REMEMBER WE HAVE A FUNCTION OF Y AND Z
NOT X AND Y LIKE WE NORMALLY WOULD
SO THESE X'S WOULD BE THESE Y'S AND THESE Y'S WOULD BE THE Z'S.
SO THE SECOND ORDER OF PARTIALS WITH RESPECT TO Y IS HERE.
THE SECOND ORDER OF PARTIAL WITH RESPECT TO Z IS HERE.
HERE'S THE MIXED PARTIAL SQUARED.
NOTICE HOW IT COMES OUT TO A POSITIVE VALUE.
SO IF D IS POSITIVE AND SO IS THE SECOND PARTIAL
WITH RESPECT TO Y LOOKING AT OUR NOTES HERE
THAT MEANS WE HAVE A RELATIVE MINIMUM AT OUR CRITICAL POINT
AND THEREFORE THESE ARE THE DIMENSIONS
THAT WOULD MINIMIZE THE COST OF OUR BOX.
THIS WAS THE X COORDINATE FROM THE PREVIOUS SLIDE.
HERE'S THE Y COORDINATE AND HERE'S THE Z COORDINATE
WHICH AGAIN ARE THE DIMENSIONS OF OUR BOX.
SO THE FRONT WIDTH WOULD BE X
WHICH IS APPROXIMATELY 4.05 INCHES.
THE DEPTH WOULD BE Y,
WHICH IS APPROXIMATELY 14.19 INCHES,
AND THE HEIGHT WOULD BE Z, WHICH IS APPROXIMATELY 6.08 INCHES.
LET'S FINISH BY LOOKING AT OUR COST FUNCTION
WHERE WE HAVE THE COST FUNCTION IN TERMS OF Y AND Z.
IN THREE DIMENSIONS THIS WOULD BE THE SURFACE
WHERE THESE LOWER AXES WOULD BE THE Y AND Z AXIS
AND THE COST WOULD BE ALONG THE VERTICAL AXIS.
WE CAN SEE THERE'S A LOW POINT HERE AND THAT OCCURRED
AT OUR CRITICAL POINT THAT WE FOUND.
I HOPE YOU FOUND THIS HELPFUL.