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AN Open up Prime RECTANGULAR BOX IS Currently being Built TO HOLD A VOLUME OF 350 CUBIC INCHES.
The bottom From the BOX IS Made out of Product COSTING 6 CENTS For every SQUARE INCH.
THE FRONT OF THE BOX Need to be DECORATED And may Price twelve CENTS For every SQUARE INCH.
THE REMAINDER OF The perimeters WILL Value 2 CENTS For every SQUARE INCH.
Discover THE DIMENSIONS THAT WILL Lessen The price of Developing THIS BOX.
LET'S Initial DIAGRAM THE BOX AS WE SEE Below WHERE The size ARE X BY Y BY Z AND BECAUSE The amount Should be 350 CUBIC INCHES We now have A CONSTRAINT THAT X x Y x Z Ought to Equivalent 350.
BUT In advance of WE Take a look at OUR Price tag Operate Allows Take a look at THE Surface area Space From the BOX.
As the Major IS OPEN, WE ONLY HAVE five FACES.
LET'S Discover the Place On the five FACES That might MAKE UP THE Surface area AREA.
See The realm From the Entrance Deal with Could well be X x Z Which might ALSO BE Similar to The world Inside the BACK And so the SURFACE AREA HAS TWO XZ Conditions.
Observe The correct Facet OR The best FACE Would've AREA Y x Z WHICH Would be the Very same AS THE Remaining.
And so the Floor Place Is made up of TWO YZ Phrases And after that Last but not least The underside HAS A location OF X x Y AND BECAUSE THE TOP IS OPEN WE ONLY HAVE A person XY Time period Within the SURFACE AREA AND NOW WE'LL CONVERT THE Area Region TO The associated fee EQUATION.
As the Base Value six CENTS PER Sq. INCH WHERE The realm OF The underside IS X x Y Observe HOW FOR THE COST FUNCTION WE MULTIPLY THE XY Expression BY six CENTS AND BECAUSE THE Entrance Expenditures 12 CENTS PER Sq. INCH The place The world From the FRONT Might be X x Z WE'LL MULTIPLY THIS XZ Expression BY twelve CENTS IN The associated fee Purpose.
THE REMAINING SIDES COST two CENTS For each Sq. INCH SO THESE A few Spots ARE ALL MULTIPLIED BY 0.
02 OR two CENTS.
COMBINING LIKE Phrases We now have THIS Price tag Operate In this article.
BUT Discover HOW We now have THREE UNKNOWNS With this EQUATION SO NOW WE'LL Make use of a CONSTRAINT TO FORM A COST EQUATION WITH TWO VARIABLES.
IF WE Resolve OUR CONSTRAINT FOR X BY DIVIDING Either side BY YZ WE May make A SUBSTITUTION FOR X INTO OUR Charge FUNCTION In which We will SUBSTITUTE THIS FRACTION In this article FOR X Below AND Right here.
IF WE DO THIS, WE GET THIS EQUATION Listed here AND IF WE SIMPLIFY Observe HOW THE Aspect OF Z SIMPLIFIES OUT AND Listed here Component OF Y SIMPLIFIES OUT.
SO FOR THIS FIRST TERM IF WE FIND THIS PRODUCT And afterwards MOVE THE Y UP WE WOULD HAVE 49Y On the -1 Then FOR THE LAST TERM IF WE FOUND THIS Solution AND MOVED THE Z UP We might HAVE + 21Z On the -1.
SO NOW OUR Objective IS TO MINIMIZE THIS Charge Functionality.
SO FOR The following Phase We will Locate the Important Details.
Crucial Factors ARE Where by THE Functionality IS GOING TO HAVE MAX OR MIN Perform VALUES And so they Happen Where by The 1st ORDER OF PARTIAL DERIVATIVES ARE Both of those EQUAL TO ZERO OR Wherever Both DOES NOT EXIST.
THEN At the time WE FIND THE Significant Factors, We are going to Figure out Irrespective of whether Now we have A MAX Or even a MIN VALUE Working with OUR Next Buy OF PARTIAL DERIVATIVES.
SO ON THIS SLIDE We are Getting The two The main Buy AND Next ORDER OF PARTIAL DERIVATIVES.
WE Need to be Somewhat CAREFUL Below While BECAUSE OUR FUNCTION Is usually a FUNCTION OF Y AND Z NOT X AND Y LIKE WE'RE Accustomed to.
SO FOR THE FIRST PARTIAL WITH RESPECT TO Y We'd DIFFERENTIATE WITH Regard TO Y TREATING Z AS A relentless WHICH WOULD GIVE US THIS PARTIAL Spinoff In this article.
FOR The primary PARTIAL WITH Regard TO Z We'd DIFFERENTIATE WITH Regard TO Z AND Handle Y AS A continuing Which might GIVE US THIS FIRST Get OF PARTIAL Spinoff.
NOW Applying THESE To start with Get OF PARTIAL DERIVATIVES WE CAN FIND THESE 2nd ORDER OF PARTIAL DERIVATIVES WHERE TO FIND The 2nd PARTIALS WITH Regard TO Y WE WOULD DIFFERENTIATE THIS PARTIAL DERIVATIVE WITH Regard TO Y All over again Offering US THIS.
The 2nd PARTIAL WITH Regard TO Z WE WOULD DIFFERENTIATE THIS PARTIAL DERIVATIVE WITH Regard TO Z Once again Supplying US THIS.
Detect The way it'S Supplied Utilizing a NEGATIVE EXPONENT AND IN FRACTION FORM Then At last With the MIXED PARTIAL OR The next Buy OF PARTIAL WITH RESPECT TO Y AND THEN Z We might DIFFERENTIATE THIS PARTIAL WITH RESPECT TO Z WHICH Discover HOW It will JUST GIVE US 0.
04.
SO NOW We will SET The 1st Purchase OF PARTIAL DERIVATIVES EQUAL TO ZERO AND Clear up Like a Process OF EQUATIONS.
SO Here i will discuss The initial ORDER OF PARTIALS SET EQUAL TO ZERO.
THIS IS A FAIRLY INVOLVED SYSTEM OF EQUATIONS WHICH We will Resolve Utilizing SUBSTITUTION.
SO I DECIDED TO Clear up The 1st EQUATION Right here FOR Z.
SO I Additional THIS TERM TO Either side In the EQUATION After which you can DIVIDED BY 0.
04 Providing US THIS Worth Right here FOR Z But when WE FIND THIS QUOTIENT AND Shift Y For the -two TO THE DENOMINATOR WE May also Create Z AS THIS FRACTION Below.
Since WE KNOW Z IS EQUAL TO THIS Portion, We could SUBSTITUTE THIS FOR Z INTO The 2nd EQUATION Right here.
Which can be WHAT WE SEE HERE BUT Observe HOW That is RAISED TO THE EXPONENT OF -two SO This may BE one, 225 Towards the -2 DIVIDED BY Y For the -4.
SO WE Will take THE RECIPROCAL WHICH WOULD GIVE US Y On the 4th DIVIDED BY 1, 500, 625 AND This is THE 21.
Since We have now AN EQUATION WITH JUST ONE VARIABLE Y We wish to Resolve THIS FOR Y.
SO FOR The initial step, You will find there's Frequent FACTOR OF Y.
SO Y = 0 WOULD SATISFY THIS EQUATION AND WOULD BE A Important Level BUT We all know WE'RE NOT Heading TO HAVE A DIMENSION OF ZERO SO We are going to JUST Overlook THAT Worth AND SET THIS EXPRESSION In this article Equivalent TO ZERO AND Address Which happens to be WHAT WE SEE In this article.
SO We'll ISOLATE THE Y CUBED TERM AND THEN CUBE ROOT Either side OF THE EQUATION.
SO IF WE Insert THIS FRACTION TO Each side Of your EQUATION And after that CHANGE THE Buy In the EQUATION This is certainly WHAT WE Would've AND NOW FROM HERE TO ISOLATE Y CUBED WE Really need to MULTIPLY With the RECIPROCAL Of the Portion HERE.
SO Discover HOW THE Remaining Facet SIMPLIFIES JUST Y CUBED AND THIS PRODUCT Here's Somewhere around THIS VALUE Listed here.
SO NOW To unravel FOR Y WE WOULD CUBE ROOT Each side From the EQUATION OR Elevate Either side OF THE EQUATION For the one/three Electricity AND This provides Y IS Close to 14.
1918, AND NOW TO FIND THE Z COORDINATE Of your Vital Issue We will USE THIS EQUATION Right here Exactly where Z = 1, 225 DIVIDED BY Y SQUARED WHICH GIVES Z IS APPROXIMATELY six.
0822.
WE DON'T Have to have IT At this moment BUT I WENT AHEAD AND FOUND THE CORRESPONDING X Price At the same time Making use of OUR VOLUME FORMULA Fix FOR X.
SO X Might be About 4.
0548.
Since WE Have only 1 CRITICAL POINT We will In all probability Think THIS Level IS GOING TO MINIMIZE THE COST Purpose BUT TO Confirm THIS WE'LL Go on and USE THE Significant Position AND The next Purchase OF PARTIAL DERIVATIVES JUST To verify.
Which means WE'LL USE THIS System In this article FOR D As well as VALUES OF THE SECOND Buy OF PARTIAL DERIVATIVES TO DETERMINE No matter whether WE HAVE A RELATIVE MAX OR MIN AT THIS Significant Issue WHEN Y IS APPROXIMATELY fourteen.
19 AND Z IS APPROXIMATELY six.
08.
HERE ARE The next ORDER OF PARTIALS THAT WE Identified Previously.
SO We are going to BE SUBSTITUTING THIS VALUE FOR Y Which Benefit FOR Z INTO THE SECOND Purchase OF PARTIALS.
WE Really should be A little bit Watchful Although Simply because Don't forget We now have A Perform OF Y AND Z NOT X AND Y LIKE WE NORMALLY WOULD SO THESE X'S Will be THESE Y'S AND THESE Y'S Could be THE Z'S.
SO The next Buy OF PARTIALS WITH RESPECT TO Y IS Right here.
The next Get OF PARTIAL WITH RESPECT TO Z IS In this article.
Here is THE MIXED PARTIAL SQUARED.
Observe HOW IT Will come OUT TO A Good Benefit.
SO IF D IS Optimistic AND SO IS The 2nd PARTIAL WITH Regard TO Y Taking a look at OUR NOTES In this article Meaning Now we have A RELATIVE Minimal AT OUR CRITICAL Issue And for that reason They are The scale That might Lower The price of OUR BOX.
THIS WAS THE X COORDINATE Within the Former SLIDE.
Here is THE Y COORDINATE AND HERE'S THE Z COORDINATE WHICH Once more ARE The scale OF OUR BOX.
Therefore the Entrance WIDTH Could well be X And that is Close to 4.
05 INCHES.
THE DEPTH Might be Y, Which happens to be APPROXIMATELY fourteen.
19 INCHES, AND The peak WOULD BE Z, Which happens to be Roughly six.
08 INCHES.
Let us End BY Investigating OUR Expense Functionality In which WE Contain the Expense Operate With regard to Y AND Z.
IN A few Proportions This may BE THE Area WHERE THESE Decreased AXES Could well be THE Y AND Z AXIS AND THE COST Could well be Alongside THE VERTICAL AXIS.
We can easily SEE THERE'S A LOW Place In this article Which OCCURRED AT OUR Significant Place THAT WE Located.
I HOPE YOU Located THIS Handy.