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AN Open up Top rated RECTANGULAR BOX IS BEING Made TO HOLD A VOLUME OF 350 CUBIC INCHES.
THE BASE Of your BOX IS Made out of Product COSTING six CENTS For every Sq. INCH.
THE Entrance OF THE BOX Need to be DECORATED AND WILL COST twelve CENTS For every SQUARE INCH.
The rest OF The perimeters WILL Charge 2 CENTS For each SQUARE INCH.
FIND THE DIMENSIONS That could Decrease THE COST OF CONSTRUCTING THIS BOX.
Let us FIRST DIAGRAM THE BOX AS WE SEE In this article In which The size ARE X BY Y BY Z AND BECAUSE The quantity MUST BE 350 CUBIC INCHES We've A CONSTRAINT THAT X x Y x Z MUST Equivalent 350.
BUT Ahead of WE Discuss OUR Price tag FUNCTION Allows TALK ABOUT THE SURFACE Region OF THE BOX.
As the Top rated IS OPEN, WE ONLY HAVE 5 FACES.
LET'S Locate the Location OF THE five FACES That may MAKE UP THE Floor Spot.
Observe The realm With the FRONT Confront Will be X x Z Which might Even be Similar to THE AREA While in the Back again SO THE Area Spot HAS TWO XZ Conditions.
Discover THE RIGHT Aspect OR THE RIGHT Facial area WOULD HAVE Place Y x Z WHICH Would be the Very same Given that the Still left.
Therefore the Floor Space Includes TWO YZ TERMS And after that FINALLY The underside HAS A location OF X x Y AND BECAUSE The highest IS Open up WE Have only One particular XY Phrase IN THE Floor Place AND NOW We will CONVERT THE Surface area AREA TO The fee EQUATION.
As the Base Price six CENTS For every SQUARE INCH The place The region OF The underside IS X x Y Recognize HOW FOR The expense Perform WE MULTIPLY THE XY Phrase BY 6 CENTS And since THE FRONT Expenses 12 CENTS PER SQUARE INCH The place The world On the Entrance Will be X x Z We will MULTIPLY THIS XZ TERM BY 12 CENTS IN THE COST FUNCTION.
THE REMAINING SIDES Charge 2 CENTS For each SQUARE INCH SO THESE A few Spots ARE ALL MULTIPLIED BY 0.
02 OR two CENTS.
COMBINING LIKE Conditions We've got THIS Price Functionality Right here.
BUT See HOW We've A few UNKNOWNS Within this EQUATION SO NOW WE'LL Utilize a CONSTRAINT TO FORM A value EQUATION WITH TWO VARIABLES.
IF WE Resolve OUR CONSTRAINT FOR X BY DIVIDING BOTH SIDES BY YZ WE CAN MAKE A SUBSTITUTION FOR X INTO OUR Price tag Functionality Exactly where WE CAN SUBSTITUTE THIS FRACTION HERE FOR X HERE AND Right here.
IF WE Try this, WE GET THIS EQUATION Below And when WE SIMPLIFY Detect HOW THE Component OF Z SIMPLIFIES OUT AND In this article Aspect OF Y SIMPLIFIES OUT.
SO FOR THIS FIRST Time period IF WE FIND THIS Merchandise Then MOVE THE Y UP WE WOULD HAVE 49Y Towards the -one And after that FOR THE LAST Phrase IF WE Located THIS Item AND MOVED THE Z UP We might HAVE + 21Z To your -one.
SO NOW OUR Aim IS TO MINIMIZE THIS Expense Perform.
SO FOR The subsequent Move We will Discover the Crucial Details.
Crucial Factors ARE In which THE Operate IS GOING TO HAVE MAX OR MIN Perform VALUES They usually Manifest Exactly where The main Buy OF PARTIAL DERIVATIVES ARE Equally Equivalent TO ZERO OR Wherever Both Isn't going to EXIST.
THEN The moment WE FIND THE Important Factors, WE'LL Establish WHETHER Now we have A MAX Or perhaps a MIN Worth Employing OUR Next Purchase OF PARTIAL DERIVATIVES.
SO ON THIS SLIDE We are Getting Both of those The very first ORDER AND Next Buy OF PARTIAL DERIVATIVES.
WE Need to be Slightly Cautious HERE While Simply because OUR Purpose Is usually a Functionality OF Y AND Z NOT X AND Y LIKE We are USED TO.
SO FOR The primary PARTIAL WITH Regard TO Y WE WOULD DIFFERENTIATE WITH RESPECT TO Y Dealing with Z AS A relentless Which might GIVE US THIS PARTIAL Spinoff Below.
FOR The primary PARTIAL WITH RESPECT TO Z WE WOULD DIFFERENTIATE WITH Regard TO Z AND Take care of Y AS A relentless WHICH WOULD GIVE US THIS FIRST ORDER OF PARTIAL Spinoff.
NOW Utilizing THESE To start with Get OF PARTIAL DERIVATIVES WE Can discover THESE SECOND Get OF PARTIAL DERIVATIVES Wherever To search out The 2nd PARTIALS WITH Regard TO Y We might DIFFERENTIATE THIS PARTIAL DERIVATIVE WITH Regard TO Y Once more GIVING US THIS.
THE SECOND PARTIAL WITH RESPECT TO Z WE WOULD DIFFERENTIATE THIS PARTIAL Spinoff WITH RESPECT TO Z AGAIN Providing US THIS.
Discover HOW IT'S Offered Employing a Adverse EXPONENT As well as in FRACTION Variety And afterwards Ultimately With the MIXED PARTIAL OR The next Purchase OF PARTIAL WITH Regard TO Y After which you can Z WE WOULD DIFFERENTIATE THIS PARTIAL WITH RESPECT TO Z WHICH Detect HOW IT WOULD JUST GIVE US 0.
04.
SO NOW WE'RE GOING TO SET The 1st Purchase OF PARTIAL DERIVATIVES Equivalent TO ZERO AND Address Like a Technique OF EQUATIONS.
SO Allow me to share The very first Buy OF PARTIALS Established Equivalent TO ZERO.
THIS Is a reasonably Included Program OF EQUATIONS WHICH We will Remedy Making use of SUBSTITUTION.
SO I DECIDED TO Remedy The main EQUATION In this article FOR Z.
SO I Extra THIS Phrase TO Either side With the EQUATION After which you can DIVIDED BY 0.
04 GIVING US THIS Worth HERE FOR Z However, if WE FIND THIS QUOTIENT AND MOVE Y For the -2 On the DENOMINATOR WE Might also Create Z AS THIS Portion Right here.
Given that WE KNOW Z IS Equivalent TO THIS Portion, WE CAN SUBSTITUTE THIS FOR Z INTO The next EQUATION Below.
And that is WHAT WE SEE Right here BUT Discover HOW That is Lifted On the EXPONENT OF -two SO THIS WOULD BE 1, 225 Towards the -2 DIVIDED BY Y To your -4.
SO WE Might take THE RECIPROCAL WHICH WOULD GIVE US Y TO THE 4th DIVIDED BY 1, five hundred, 625 AND Here is THE 21.
Given that We've got AN EQUATION WITH JUST ONE VARIABLE Y We wish to Address THIS FOR Y.
SO FOR THE FIRST STEP, You will find there's Popular Issue OF Y.
SO Y = 0 WOULD Fulfill THIS EQUATION AND Will be A Important Stage BUT WE KNOW We are NOT GOING To possess a DIMENSION OF ZERO SO WE'LL JUST Dismiss THAT VALUE AND Established THIS EXPRESSION Right here EQUAL TO ZERO AND SOLVE Which can be WHAT WE SEE Listed here.
SO We will ISOLATE THE Y CUBED Phrase And afterwards CUBE ROOT BOTH SIDES In the EQUATION.
SO IF WE Insert THIS Portion TO Each side From the EQUATION After which CHANGE THE Buy With the EQUATION This is certainly WHAT WE WOULD HAVE AND NOW FROM Right here TO ISOLATE Y CUBED WE Should MULTIPLY Through the RECIPROCAL Of the FRACTION Right here.
SO Discover HOW THE Still left Facet SIMPLIFIES JUST Y CUBED Which Merchandise HERE IS Roughly THIS VALUE Listed here.
SO NOW To resolve FOR Y We'd CUBE ROOT Each side OF THE EQUATION OR RAISE Each side OF THE EQUATION On the one/three Electricity AND THIS GIVES Y IS Around fourteen.
1918, AND NOW TO Locate the Z COORDINATE From the CRITICAL Level We can easily USE THIS EQUATION Right here Where by Z = one, 225 DIVIDED BY Y SQUARED WHICH GIVES Z IS Close to six.
0822.
WE DON'T Require IT Today BUT I WENT Forward AND FOUND THE CORRESPONDING X Worth Also Working with OUR Quantity Components Resolve FOR X.
SO X Could be Somewhere around four.
0548.
Mainly because WE Have only Just one Crucial Position We could Most likely Think THIS Issue Will probably Decrease The price Functionality BUT TO VERIFY THIS WE'LL Go on and Utilize the Crucial Position AND The next ORDER OF PARTIAL DERIVATIVES JUST To make certain.
Which means WE'LL USE THIS Formulation Right here FOR D As well as the VALUES OF THE SECOND Buy OF PARTIAL DERIVATIVES TO DETERMINE No matter if WE HAVE A RELATIVE MAX OR MIN AT THIS Vital Level WHEN Y IS Roughly fourteen.
19 AND Z IS APPROXIMATELY 6.
08.
Allow me to share THE SECOND ORDER OF PARTIALS THAT WE FOUND EARLIER.
SO We are going to BE SUBSTITUTING THIS VALUE FOR Y Which Worth FOR Z INTO THE SECOND Buy OF PARTIALS.
WE Needs to be A bit Very careful However Simply because Recall We've A Perform OF Y AND Z NOT X AND Y LIKE WE Generally WOULD SO THESE X'S Could well be THESE Y'S AND THESE Y'S Could be THE Z'S.
SO THE SECOND ORDER OF PARTIALS WITH RESPECT TO Y IS In this article.
THE SECOND ORDER OF PARTIAL WITH Regard TO Z IS Listed here.
This is THE MIXED PARTIAL SQUARED.
Recognize HOW IT Arrives OUT To some Good VALUE.
SO IF D IS Optimistic AND SO IS THE SECOND PARTIAL WITH Regard TO Y Investigating OUR NOTES Right here Which means We've A RELATIVE Least AT OUR Important Position And thus These are typically The size That might Lessen The price of OUR BOX.
THIS WAS THE X COORDINATE In the PREVIOUS SLIDE.
HERE'S THE Y COORDINATE AND Here is THE Z COORDINATE WHICH Once more ARE The size OF OUR BOX.
And so the Entrance WIDTH Could well be X WHICH IS Around four.
05 INCHES.
THE DEPTH Could well be Y, Which can be Roughly fourteen.
19 INCHES, AND The peak WOULD BE Z, Which happens to be Around 6.
08 INCHES.
LET'S Complete BY Checking out OUR Price FUNCTION In which WE Provide the Charge Functionality IN TERMS OF Y AND Z.
IN A few Proportions This is able to BE THE SURFACE Exactly where THESE LOWER AXES Will be THE Y AND Z AXIS AND The expense Will be ALONG THE VERTICAL AXIS.
We could SEE There is a LOW POINT Right here Which OCCURRED AT OUR Crucial Level THAT WE Uncovered.
I HOPE YOU Uncovered THIS Valuable.