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AN OPEN Prime RECTANGULAR BOX IS Getting Created TO HOLD A Quantity OF 350 CUBIC INCHES.
THE BASE With the BOX IS Created from Substance COSTING six CENTS For every SQUARE INCH.
THE Entrance OF THE BOX Should be DECORATED And can Value 12 CENTS PER Sq. INCH.
THE REMAINDER OF The edges WILL Charge 2 CENTS For each SQUARE INCH.
Obtain THE DIMENSIONS That can Lessen The price of Setting up THIS BOX.
LET'S Very first DIAGRAM THE BOX AS WE SEE Right here Where by The size ARE X BY Y BY Z AND BECAUSE THE VOLUME Have to be 350 CUBIC INCHES We have now A CONSTRAINT THAT X x Y x Z Should Equivalent 350.
BUT Prior to WE Take a look at OUR COST Functionality LETS Speak about THE Floor Spot OF THE BOX.
As the Leading IS OPEN, WE Have only five FACES.
LET'S Locate the Spot On the five FACES That may MAKE UP THE SURFACE Region.
Recognize The region On the Entrance Encounter Can be X x Z WHICH WOULD ALSO BE THE SAME AS The region During the BACK Hence the Surface area Spot HAS TWO XZ TERMS.
Observe The ideal Facet OR The ideal Confront Might have Region Y x Z WHICH WILL BE THE Identical Because the Still left.
Hence the Floor AREA Consists of TWO YZ TERMS After which FINALLY THE BOTTOM HAS AN AREA OF X x Y AND BECAUSE The very best IS OPEN WE ONLY HAVE 1 XY TERM Inside the Surface area Region AND NOW We are going to CONVERT THE Area AREA TO The price EQUATION.
As the Base COST 6 CENTS For every Sq. INCH The place The realm OF The underside IS X x Y See HOW FOR THE COST Purpose WE MULTIPLY THE XY Phrase BY 6 CENTS And since THE Entrance Expenditures twelve CENTS For every SQUARE INCH Where by THE AREA With the Entrance Can be X x Z We will MULTIPLY THIS XZ TERM BY twelve CENTS IN The expense Purpose.
THE REMAINING SIDES Price two CENTS For each SQUARE INCH SO THESE THREE Regions ARE ALL MULTIPLIED BY 0.
02 OR 2 CENTS.
COMBINING LIKE TERMS We've THIS Value Purpose In this article.
BUT See HOW WE HAVE THREE UNKNOWNS In this particular EQUATION SO NOW We are going to USE A CONSTRAINT TO Variety A price EQUATION WITH TWO VARIABLES.
IF WE Address OUR CONSTRAINT FOR X BY DIVIDING Each side BY YZ WE Will make A SUBSTITUTION FOR X INTO OUR Price tag Perform Wherever We can easily SUBSTITUTE THIS FRACTION Listed here FOR X Right here AND Below.
IF WE Make this happen, WE GET THIS EQUATION Listed here And when WE SIMPLIFY Observe HOW THE Aspect OF Z SIMPLIFIES OUT AND Right here Variable OF Y SIMPLIFIES OUT.
SO FOR THIS FIRST TERM IF WE FIND THIS Merchandise And after that Transfer THE Y UP WE Might have 49Y TO THE -1 Then FOR The final TERM IF WE Located THIS PRODUCT AND MOVED THE Z UP We might HAVE + 21Z TO THE -one.
SO NOW OUR Intention IS To attenuate THIS Value Functionality.
SO FOR The following Phase We are going to Discover the Significant Details.
Important Factors ARE Where by THE Operate Will almost certainly HAVE MAX OR MIN Perform VALUES They usually Manifest Where by The initial ORDER OF PARTIAL DERIVATIVES ARE Each EQUAL TO ZERO OR Exactly where EITHER Will not EXIST.
THEN When WE Locate the Vital Details, We will DETERMINE Regardless of whether WE HAVE A MAX Or maybe a MIN Worth Working with OUR Next Buy OF PARTIAL DERIVATIVES.
SO ON THIS SLIDE WE'RE Obtaining BOTH The initial Buy AND Next Get OF PARTIAL DERIVATIVES.
WE Need to be A LITTLE Very careful In this article While Simply because OUR Functionality Can be a Operate OF Y AND Z NOT X AND Y LIKE We are Accustomed to.
SO FOR The initial PARTIAL WITH Regard TO Y WE WOULD DIFFERENTIATE WITH RESPECT TO Y TREATING Z AS A CONSTANT WHICH WOULD GIVE US THIS PARTIAL DERIVATIVE Listed here.
FOR The main PARTIAL WITH Regard TO Z We might DIFFERENTIATE WITH RESPECT TO Z AND Handle Y AS A relentless WHICH WOULD GIVE US THIS FIRST Get OF PARTIAL DERIVATIVE.
NOW Working with THESE To start with Purchase OF PARTIAL DERIVATIVES WE Can discover THESE Next Purchase OF PARTIAL DERIVATIVES Wherever To seek out THE SECOND PARTIALS WITH RESPECT TO Y WE WOULD DIFFERENTIATE THIS PARTIAL Spinoff WITH Regard TO Y AGAIN GIVING US THIS.
The next PARTIAL WITH RESPECT TO Z We'd DIFFERENTIATE THIS PARTIAL DERIVATIVE WITH Regard TO Z All over again GIVING US THIS.
Discover HOW IT'S Specified Utilizing a Unfavorable EXPONENT As well as in FRACTION Sort After which you can Ultimately For that Combined PARTIAL OR The 2nd Purchase OF PARTIAL WITH RESPECT TO Y AND THEN Z WE WOULD DIFFERENTIATE THIS PARTIAL WITH Regard TO Z WHICH Discover HOW It might JUST GIVE US 0.
04.
SO NOW We'll Established The very first ORDER OF PARTIAL DERIVATIVES EQUAL TO ZERO AND Clear up AS A Procedure OF EQUATIONS.
SO Allow me to share The primary ORDER OF PARTIALS SET EQUAL TO ZERO.
THIS Is a reasonably Included Procedure OF EQUATIONS WHICH We are going to Remedy USING SUBSTITUTION.
SO I DECIDED TO SOLVE The initial EQUATION Right here FOR Z.
SO I Included THIS TERM TO BOTH SIDES From the EQUATION And afterwards DIVIDED BY 0.
04 Providing US THIS VALUE Listed here FOR Z But when WE FIND THIS QUOTIENT AND Go Y Towards the -2 TO THE DENOMINATOR WE Might also WRITE Z AS THIS Portion Listed here.
Since WE KNOW Z IS EQUAL TO THIS FRACTION, We are able to SUBSTITUTE THIS FOR Z INTO THE SECOND EQUATION Right here.
And that is WHAT WE SEE HERE BUT Recognize HOW That is Lifted TO THE EXPONENT OF -2 SO This may BE one, 225 For the -2 DIVIDED BY Y Towards the -4.
SO WE Might take THE RECIPROCAL Which might GIVE US Y For the 4th DIVIDED BY 1, five hundred, 625 AND HERE'S THE 21.
Since We've AN EQUATION WITH JUST ONE VARIABLE Y We wish to Address THIS FOR Y.
SO FOR Step one, There exists a Prevalent Component OF Y.
SO Y = 0 WOULD SATISFY THIS EQUATION AND WOULD BE A Essential Position BUT We all know We are NOT GOING TO HAVE A DIMENSION OF ZERO SO We will JUST Overlook THAT Price AND SET THIS EXPRESSION Right here Equivalent TO ZERO AND Clear up And that is WHAT WE SEE Listed here.
SO We'll ISOLATE THE Y CUBED TERM AND THEN Dice ROOT Each side From the EQUATION.
Therefore if WE Include THIS Portion TO Either side From the EQUATION And after that Alter the Purchase In the EQUATION This can be WHAT WE Might have AND NOW FROM Listed here TO ISOLATE Y CUBED WE Must MULTIPLY Because of the RECIPROCAL Of the FRACTION Listed here.
SO Recognize HOW THE LEFT Facet SIMPLIFIES JUST Y CUBED AND THIS Products Here's Around THIS Price Listed here.
SO NOW TO SOLVE FOR Y We'd CUBE ROOT BOTH SIDES On the EQUATION OR Elevate Each side Of your EQUATION Into the 1/3 Ability AND This offers Y IS APPROXIMATELY 14.
1918, AND NOW TO Discover the Z COORDINATE From the Significant Position We are able to USE THIS EQUATION Right here Where by Z = one, 225 DIVIDED BY Y SQUARED Which provides Z IS Somewhere around six.
0822.
We do not Will need IT At this moment BUT I WENT Forward AND FOUND THE CORRESPONDING X VALUE Also Working with OUR VOLUME FORMULA Clear up FOR X.
SO X WOULD BE Somewhere around four.
0548.
For the reason that WE Have only Just one Crucial POINT We will Possibly Presume THIS Place Will almost certainly Limit THE COST Functionality BUT TO Confirm THIS WE'LL GO AHEAD AND Make use of the CRITICAL Position AND The 2nd ORDER OF PARTIAL DERIVATIVES JUST To be certain.
MEANING WE'LL USE THIS Components HERE FOR D Plus the VALUES OF The 2nd Get OF PARTIAL DERIVATIVES TO DETERMINE No matter if WE HAVE A RELATIVE MAX OR MIN AT THIS CRITICAL Place WHEN Y IS Around fourteen.
19 AND Z IS APPROXIMATELY six.
08.
Listed below are THE SECOND ORDER OF PARTIALS THAT WE Discovered EARLIER.
SO WE'LL BE SUBSTITUTING THIS Benefit FOR Y AND THIS Worth FOR Z INTO The 2nd Buy OF PARTIALS.
WE Really should be Somewhat Thorough However BECAUSE Don't forget WE HAVE A Operate OF Y AND Z NOT X AND Y LIKE WE Generally WOULD SO THESE X'S Will be THESE Y'S AND THESE Y'S Could well be THE Z'S.
SO The 2nd Buy OF PARTIALS WITH RESPECT TO Y IS Below.
The 2nd Buy OF PARTIAL WITH Regard TO Z IS HERE.
HERE'S THE Combined PARTIAL SQUARED.
NOTICE The way it Arrives OUT Into a Constructive Benefit.
Therefore if D IS Favourable AND SO IS The next PARTIAL WITH RESPECT TO Y Investigating OUR NOTES In this article THAT MEANS We've A RELATIVE Bare minimum AT OUR CRITICAL Level AND THEREFORE THESE ARE THE DIMENSIONS THAT WOULD Lessen The price of OUR BOX.
THIS WAS THE X COORDINATE From your Prior SLIDE.
Here is THE Y COORDINATE AND This is THE Z COORDINATE WHICH Once again ARE THE DIMENSIONS OF OUR BOX.
SO THE FRONT WIDTH WOULD BE X That's APPROXIMATELY four.
05 INCHES.
THE DEPTH Could well be Y, WHICH IS APPROXIMATELY 14.
19 INCHES, AND The peak Will be Z, That's Roughly six.
08 INCHES.
LET'S Complete BY Considering OUR COST FUNCTION Where by WE HAVE THE Price tag Functionality Concerning Y AND Z.
IN THREE Proportions This could BE THE SURFACE WHERE THESE Reduce AXES Could well be THE Y AND Z AXIS AND The price Could well be ALONG THE VERTICAL AXIS.
WE CAN SEE There is a Reduced POINT Listed here Which Happened AT OUR Crucial Place THAT WE Located.
I HOPE YOU Observed THIS Valuable.