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AN OPEN Major RECTANGULAR BOX IS Getting Manufactured TO HOLD A Quantity OF 350 CUBIC INCHES.
The bottom Of your BOX IS Made out of Substance COSTING six CENTS For each Sq. INCH.
THE Entrance From the BOX Need to be DECORATED And may Price 12 CENTS For each SQUARE INCH.
The rest OF The edges WILL Price two CENTS For each Sq. INCH.
Discover The size That can Reduce The expense of CONSTRUCTING THIS BOX.
Let us FIRST DIAGRAM THE BOX AS WE SEE Below The place The scale ARE X BY Y BY Z AND BECAUSE The quantity Need to be 350 CUBIC INCHES We've A CONSTRAINT THAT X x Y x Z Should EQUAL 350.
BUT Right before WE Look at OUR Price Functionality LETS Take a look at THE SURFACE Region On the BOX.
BECAUSE THE Leading IS Open up, WE Have only 5 FACES.
LET'S Locate the Place On the 5 FACES THAT WOULD MAKE UP THE Surface area Region.
Recognize The region On the Entrance Encounter Could be X x Z Which might ALSO BE Similar to THE AREA While in the Again Therefore the SURFACE Location HAS TWO XZ Conditions.
See The ideal Aspect OR THE RIGHT Encounter WOULD HAVE Spot Y x Z WHICH Would be the Similar Since the Still left.
Hence the SURFACE AREA Consists of TWO YZ TERMS And afterwards Eventually THE BOTTOM HAS AN AREA OF X x Y AND BECAUSE The highest IS Open up WE Have only One particular XY TERM IN THE SURFACE Space AND NOW WE'LL Transform THE SURFACE Location TO The expense EQUATION.
Since the BOTTOM Charge 6 CENTS For every SQUARE INCH Exactly where The region OF The underside IS X x Y Observe HOW FOR The associated fee FUNCTION WE MULTIPLY THE XY TERM BY six CENTS And since THE FRONT Expenses twelve CENTS For every SQUARE INCH The place The realm From the Entrance Will be X x Z We are going to MULTIPLY THIS XZ TERM BY twelve CENTS IN The associated fee Purpose.
THE REMAINING SIDES Charge two CENTS PER SQUARE INCH SO THESE 3 AREAS ARE ALL MULTIPLIED BY 0.
02 OR 2 CENTS.
COMBINING LIKE TERMS We have now THIS Price Functionality Listed here.
BUT See HOW WE HAVE THREE UNKNOWNS On this EQUATION SO NOW We are going to USE A CONSTRAINT TO FORM A value EQUATION WITH TWO VARIABLES.
IF WE Address OUR CONSTRAINT FOR X BY DIVIDING Either side BY YZ WE Could make A SUBSTITUTION FOR X INTO OUR Price Perform Exactly where We can easily SUBSTITUTE THIS FRACTION In this article FOR X In this article AND HERE.
IF WE Try this, WE GET THIS EQUATION Right here AND IF WE SIMPLIFY See HOW THE FACTOR OF Z SIMPLIFIES OUT AND Below FACTOR OF Y SIMPLIFIES OUT.
SO FOR THIS FIRST Time period IF WE FIND THIS Products AND THEN Go THE Y UP WE WOULD HAVE 49Y Into the -1 And after that FOR THE LAST Time period IF WE FOUND THIS Product or service AND MOVED THE Z UP We would HAVE + 21Z TO THE -one.
SO NOW OUR Target IS TO MINIMIZE THIS COST Functionality.
SO FOR THE NEXT Stage We are going to Discover the Important POINTS.
Crucial POINTS ARE Where by THE FUNCTION IS GOING TO HAVE MAX OR MIN Purpose VALUES They usually Manifest Exactly where The initial Get OF PARTIAL DERIVATIVES ARE Both of those Equivalent TO ZERO OR Where by Both Would not EXIST.
THEN The moment WE FIND THE Significant Details, WE'LL DETERMINE Whether or not We've A MAX OR A MIN Price Working with OUR SECOND ORDER OF PARTIAL DERIVATIVES.
SO ON THIS SLIDE We are Discovering Equally The very first Purchase AND SECOND Buy OF PARTIAL DERIVATIVES.
WE Must be A LITTLE CAREFUL Right here THOUGH Mainly because OUR Perform Can be a Functionality OF Y AND Z NOT X AND Y LIKE We are Accustomed to.
SO FOR THE FIRST PARTIAL WITH Regard TO Y WE WOULD DIFFERENTIATE WITH Regard TO Y Managing Z AS A continuing Which might GIVE US THIS PARTIAL Spinoff HERE.
FOR The main PARTIAL WITH Regard TO Z WE WOULD DIFFERENTIATE WITH RESPECT TO Z AND TREAT Y AS A relentless Which might GIVE US This primary ORDER OF PARTIAL By-product.
NOW Working with THESE Very first Get OF PARTIAL DERIVATIVES WE Can discover THESE 2nd Buy OF PARTIAL DERIVATIVES Where by To search out The 2nd PARTIALS WITH RESPECT TO Y We might DIFFERENTIATE THIS PARTIAL By-product WITH RESPECT TO Y Once more GIVING US THIS.
The next PARTIAL WITH Regard TO Z We might DIFFERENTIATE THIS PARTIAL DERIVATIVE WITH RESPECT TO Z AGAIN Offering US THIS.
Recognize HOW IT'S Supplied Employing a Destructive EXPONENT As well as in FRACTION Type After which you can Eventually To the Combined PARTIAL OR THE SECOND Purchase OF PARTIAL WITH RESPECT TO Y After which you can Z We might DIFFERENTIATE THIS PARTIAL WITH Regard TO Z WHICH See HOW It will JUST GIVE US 0.
04.
SO NOW We'll Established The very first Purchase OF PARTIAL DERIVATIVES EQUAL TO ZERO AND Address Like a Program OF EQUATIONS.
SO Listed here are The main Buy OF PARTIALS Established EQUAL TO ZERO.
THIS IS A FAIRLY Associated Procedure OF EQUATIONS WHICH We are going to Clear up Working with SUBSTITUTION.
SO I Chose to Remedy The 1st EQUATION Right here FOR Z.
SO I ADDED THIS TERM TO Either side OF THE EQUATION After which you can DIVIDED BY 0.
04 Providing US THIS Price Right here FOR Z BUT IF WE FIND THIS QUOTIENT AND Transfer Y On the -two Into the DENOMINATOR WE May also Create Z AS THIS FRACTION HERE.
Given that We all know Z IS EQUAL TO THIS FRACTION, We will SUBSTITUTE THIS FOR Z INTO The next EQUATION In this article.
Which can be WHAT WE SEE Listed here BUT NOTICE HOW This is often RAISED For the EXPONENT OF -2 SO THIS WOULD BE 1, 225 TO THE -2 DIVIDED BY Y To your -four.
SO WE Will take THE RECIPROCAL WHICH WOULD GIVE US Y On the 4th DIVIDED BY one, 500, 625 AND Here is THE 21.
Given that We now have AN EQUATION WITH Only one VARIABLE Y We wish to Fix THIS FOR Y.
SO FOR Step one, You will find a Typical Issue OF Y.
SO Y = 0 WOULD Fulfill THIS EQUATION AND Could be A Significant Position BUT WE KNOW WE'RE NOT Heading To possess a DIMENSION OF ZERO SO We are going to JUST Disregard THAT VALUE AND Established THIS EXPRESSION HERE Equivalent TO ZERO AND Clear up And that is WHAT WE SEE HERE.
SO We will ISOLATE THE Y CUBED Phrase And afterwards Dice ROOT Each side In the EQUATION.
Therefore if WE Increase THIS Portion TO Each side In the EQUATION After which CHANGE THE Purchase In the EQUATION This really is WHAT WE Would've AND NOW FROM Listed here TO ISOLATE Y CUBED WE Really have to MULTIPLY From the RECIPROCAL Of the FRACTION Listed here.
SO NOTICE HOW THE LEFT Facet SIMPLIFIES JUST Y CUBED AND THIS Item Here's About THIS VALUE HERE.
SO NOW TO SOLVE FOR Y We'd Dice ROOT Each side On the EQUATION OR Elevate BOTH SIDES With the EQUATION TO THE one/three Ability AND THIS GIVES Y IS Around 14.
1918, AND NOW TO Locate the Z COORDINATE With the Essential Level WE CAN USE THIS EQUATION In this article The place Z = 1, 225 DIVIDED BY Y SQUARED WHICH GIVES Z IS Somewhere around six.
0822.
WE DON'T Need to have IT At this time BUT I WENT Forward AND FOUND THE CORRESPONDING X VALUE Also Utilizing OUR VOLUME Components Fix FOR X.
SO X Could well be Somewhere around four.
0548.
Simply because WE ONLY HAVE ONE Significant Position We could PROBABLY Think THIS Position Will Lessen THE COST FUNCTION BUT TO Validate THIS WE'LL Go on and USE THE Essential Position AND The 2nd Purchase OF PARTIAL DERIVATIVES JUST To be sure.
Indicating We will USE THIS Method Below FOR D Along with the VALUES OF The 2nd Purchase OF PARTIAL DERIVATIVES To find out Whether or not We have now A RELATIVE MAX OR MIN AT THIS Significant Position WHEN Y IS Close to fourteen.
19 AND Z IS Around 6.
08.
Allow me to share THE SECOND Buy OF PARTIALS THAT WE Located EARLIER.
SO We are going to BE SUBSTITUTING THIS Benefit FOR Y AND THIS Benefit FOR Z INTO The next ORDER OF PARTIALS.
WE Needs to be Somewhat Very careful Even though Mainly because Bear in mind We now have A Perform 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 SECOND Purchase OF PARTIALS WITH Regard TO Y IS Right here.
THE SECOND Get OF PARTIAL WITH RESPECT TO Z IS Below.
HERE'S THE MIXED PARTIAL SQUARED.
Discover The way it COMES OUT Into a Good Benefit.
Therefore if D IS Favourable AND SO IS The 2nd PARTIAL WITH Regard TO Y Thinking about OUR NOTES Right here Which means Now we have A RELATIVE Least AT OUR Crucial Stage AND THEREFORE THESE ARE The scale That might Lower The expense of OUR BOX.
THIS WAS THE X COORDINATE Through the PREVIOUS SLIDE.
HERE'S THE Y COORDINATE AND HERE'S THE Z COORDINATE WHICH Yet again ARE The scale OF OUR BOX.
And so the Entrance WIDTH Could be X Which can be Close to 4.
05 INCHES.
THE DEPTH Might be Y, Which happens to be Close to 14.
19 INCHES, AND The peak Could well be Z, Which can be Close to six.
08 INCHES.
LET'S Complete BY Thinking about OUR Value Operate Where by WE Provide the COST Operate Regarding Y AND Z.
IN A few DIMENSIONS This is able to BE THE Surface area Exactly where THESE Lessen AXES Might be THE Y AND Z AXIS AND The associated fee Will be Together THE VERTICAL AXIS.
We could SEE There is a Small Position Below AND THAT OCCURRED AT OUR Vital Place THAT WE Located.
I HOPE YOU Identified THIS HELPFUL.