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AN Open up Major RECTANGULAR BOX IS Remaining Produced To carry A Quantity OF 350 CUBIC INCHES.
THE BASE From the BOX IS Created from Product COSTING 6 CENTS For every SQUARE INCH.
THE FRONT On the BOX Need to be DECORATED And may Expense twelve CENTS For every SQUARE INCH.
THE REMAINDER OF The edges WILL Expense 2 CENTS For each Sq. INCH.
Obtain THE DIMENSIONS Which will Lower The price of CONSTRUCTING THIS BOX.
LET'S To start with DIAGRAM THE BOX AS WE SEE HERE WHERE The scale ARE X BY Y BY Z And since The quantity Needs to be 350 CUBIC INCHES We've A CONSTRAINT THAT X x Y x Z Have to EQUAL 350.
BUT Right before WE Discuss OUR Charge Perform Allows Take a look at THE Floor Place In the BOX.
As the Major IS Open up, WE Have only 5 FACES.
Let us Discover the Spot From the five FACES THAT WOULD MAKE UP THE SURFACE Spot.
Detect The world In the Entrance Experience Could be X x Z Which might Even be THE SAME AS The region While in the Back again Hence the Area Spot HAS TWO XZ Phrases.
Discover The correct Aspect OR THE RIGHT FACE Might have Location Y x Z WHICH WILL BE THE SAME Since the Remaining.
Therefore the Surface area Spot Consists of TWO YZ Phrases After which Lastly The underside HAS AN AREA OF X x Y And since THE TOP IS OPEN WE ONLY HAVE ONE XY Expression While in the SURFACE AREA AND NOW We will Transform THE Floor Space TO The expense EQUATION.
As the Base Expense six CENTS For each SQUARE INCH Exactly where The realm OF THE BOTTOM IS X x Y NOTICE HOW FOR The price Perform WE MULTIPLY THE XY Phrase BY 6 CENTS And since THE FRONT COSTS 12 CENTS PER SQUARE INCH Wherever THE AREA OF THE FRONT WOULD BE X x Z We will MULTIPLY THIS XZ Time period BY 12 CENTS IN THE COST Purpose.
THE REMAINING SIDES Value 2 CENTS For each SQUARE INCH SO THESE A few Regions ARE ALL MULTIPLIED BY 0.
02 OR two CENTS.
COMBINING LIKE Phrases We now have THIS Expense Functionality Below.
BUT Detect HOW We've THREE UNKNOWNS During this EQUATION SO NOW We will USE A CONSTRAINT TO Sort A price EQUATION WITH TWO VARIABLES.
IF WE Remedy OUR CONSTRAINT FOR X BY DIVIDING Each side BY YZ WE Might make A SUBSTITUTION FOR X INTO OUR Expense Perform Wherever We will SUBSTITUTE THIS FRACTION In this article FOR X HERE AND Right here.
IF WE DO THIS, WE GET THIS EQUATION HERE And when WE SIMPLIFY Recognize HOW THE FACTOR OF Z SIMPLIFIES OUT AND Below Variable OF Y SIMPLIFIES OUT.
SO FOR This primary Time period IF WE FIND THIS Product or service And afterwards Shift THE Y UP WE Might have 49Y TO THE -1 And afterwards FOR THE LAST TERM IF WE FOUND THIS Product or service AND MOVED THE Z UP We would HAVE + 21Z Towards the -1.
SO NOW OUR GOAL IS To attenuate THIS COST FUNCTION.
SO FOR Another STEP We are going to Discover the Essential POINTS.
Important Factors ARE Where by THE Functionality Will almost certainly HAVE MAX OR MIN FUNCTION VALUES And so they Take place Wherever The primary Buy OF PARTIAL DERIVATIVES ARE Both of those Equivalent TO ZERO OR Where by EITHER Will not EXIST.
THEN The moment WE FIND THE Important POINTS, We will Ascertain Irrespective of whether We now have A MAX Or perhaps a MIN Worth Working with OUR Next Get OF PARTIAL DERIVATIVES.
SO ON THIS SLIDE We are Obtaining The two The very first Get AND Next Buy OF PARTIAL DERIVATIVES.
WE Must be A little bit Cautious HERE However Since OUR Purpose Can be a Operate OF Y AND Z NOT X AND Y LIKE WE'RE USED TO.
SO FOR The very first PARTIAL WITH RESPECT TO Y We'd DIFFERENTIATE WITH RESPECT TO Y Managing Z AS A continuing WHICH WOULD GIVE US THIS PARTIAL By-product HERE.
FOR The initial PARTIAL WITH RESPECT TO Z We'd DIFFERENTIATE WITH RESPECT TO Z AND Handle Y AS A continuing WHICH WOULD GIVE US This primary ORDER OF PARTIAL By-product.
NOW Working with THESE Initially ORDER OF PARTIAL DERIVATIVES WE CAN FIND THESE Next Get OF PARTIAL DERIVATIVES Wherever To search out The next PARTIALS WITH RESPECT TO Y WE WOULD DIFFERENTIATE THIS PARTIAL DERIVATIVE WITH RESPECT TO Y AGAIN Providing US THIS.
The next PARTIAL WITH Regard TO Z We'd DIFFERENTIATE THIS PARTIAL By-product WITH Regard TO Z All over again Supplying US THIS.
Recognize The way it'S Offered USING A Unfavorable EXPONENT As well as in Portion FORM After which you can Last but not least To the MIXED PARTIAL OR The next Purchase OF PARTIAL WITH RESPECT TO Y And afterwards Z We'd DIFFERENTIATE THIS PARTIAL WITH RESPECT TO Z WHICH See HOW It could JUST GIVE US 0.
04.
SO NOW We'll SET THE FIRST ORDER OF PARTIAL DERIVATIVES EQUAL TO ZERO AND Fix AS A Procedure OF EQUATIONS.
SO Here i will discuss THE FIRST Purchase OF PARTIALS SET Equivalent TO ZERO.
THIS Is a reasonably Concerned Program OF EQUATIONS WHICH WE'LL Address Employing SUBSTITUTION.
SO I Made a decision to SOLVE The initial EQUATION Right here FOR Z.
SO I Additional THIS Time period TO BOTH SIDES On the EQUATION After which you can DIVIDED BY 0.
04 Providing US THIS Benefit In this article FOR Z But when We discover THIS QUOTIENT AND MOVE Y Into the -2 On the DENOMINATOR WE CAN ALSO Generate Z AS THIS Portion Below.
Given that WE KNOW Z IS Equivalent TO THIS FRACTION, We are able to SUBSTITUTE THIS FOR Z INTO The next EQUATION Below.
Which happens to be WHAT WE SEE Below BUT Observe HOW That is Lifted Into the EXPONENT OF -two SO This might BE one, 225 On the -2 DIVIDED BY Y TO THE -four.
SO WE CAN TAKE THE RECIPROCAL Which might GIVE US Y Into the 4th DIVIDED BY 1, 500, 625 AND HERE'S THE 21.
Since WE HAVE AN EQUATION WITH JUST ONE VARIABLE Y We wish to SOLVE THIS FOR Y.
SO FOR THE FIRST STEP, You will find a Typical Element OF Y.
SO Y = 0 WOULD SATISFY THIS EQUATION AND Will be A CRITICAL Level BUT We all know We are NOT GOING To possess a DIMENSION OF ZERO SO We will JUST IGNORE THAT Price AND SET THIS EXPRESSION Right here EQUAL TO ZERO AND Resolve Which happens to be WHAT WE SEE Below.
SO We'll ISOLATE THE Y CUBED Time period Then CUBE ROOT BOTH SIDES With the EQUATION.
SO IF WE ADD THIS Portion TO Each side With the EQUATION After which you can CHANGE THE Buy Of your EQUATION This is certainly WHAT WE WOULD HAVE AND NOW FROM Right here TO ISOLATE Y CUBED WE Must MULTIPLY With the RECIPROCAL OF THIS FRACTION HERE.
SO NOTICE HOW THE Remaining Aspect SIMPLIFIES JUST Y CUBED AND THIS Solution HERE IS Roughly THIS Price In this article.
SO NOW To unravel FOR Y We'd CUBE ROOT Either side With the EQUATION OR RAISE Either side With the EQUATION Into the 1/three Energy AND THIS GIVES Y IS Somewhere around 14.
1918, AND NOW TO Discover the Z COORDINATE With the Crucial POINT WE CAN USE THIS EQUATION In this article In which Z = one, 225 DIVIDED BY Y SQUARED Which provides Z IS Somewhere around 6.
0822.
WE DON'T Will need IT Right this moment BUT I WENT In advance And located THE CORRESPONDING X Benefit In addition Applying OUR VOLUME FORMULA Remedy FOR X.
SO X Could well be APPROXIMATELY 4.
0548.
Simply because WE Have only Just one Important Place We can easily Likely ASSUME THIS Level Will probably Lessen The expense FUNCTION BUT TO Confirm THIS WE'LL GO AHEAD AND Utilize the Important Stage AND THE SECOND Get OF PARTIAL DERIVATIVES JUST To ensure.
MEANING We are going to USE THIS Formulation In this article FOR D And also the VALUES OF The next Purchase OF PARTIAL DERIVATIVES To ascertain WHETHER We have now A RELATIVE MAX OR MIN AT THIS Crucial POINT WHEN Y IS Somewhere around 14.
19 AND Z IS Somewhere around six.
08.
Here's The 2nd Buy OF PARTIALS THAT WE FOUND EARLIER.
SO WE'LL BE SUBSTITUTING THIS VALUE FOR Y AND THIS Price FOR Z INTO The next Get OF PARTIALS.
WE Need to be Just a little Thorough While Mainly because REMEMBER WE HAVE A Purpose OF Y AND Z NOT X AND Y LIKE WE Commonly WOULD SO THESE X'S Can be THESE Y'S AND THESE Y'S Might be THE Z'S.
SO The next Purchase OF PARTIALS WITH Regard TO Y IS Listed here.
The next Get OF PARTIAL WITH Regard TO Z IS HERE.
This is THE Blended PARTIAL SQUARED.
Observe The way it COMES OUT To some Constructive Benefit.
Therefore if D IS Favourable AND SO IS The 2nd PARTIAL WITH RESPECT TO Y Taking a look at OUR NOTES In this article Which means We now have A RELATIVE Least AT OUR Vital Issue And so These are definitely The scale That might MINIMIZE THE COST OF OUR BOX.
THIS WAS THE X COORDINATE Within the Earlier SLIDE.
HERE'S THE Y COORDINATE AND Here is THE Z COORDINATE WHICH All over again ARE The size OF OUR BOX.
Therefore the Entrance WIDTH Can be X Which can be Close to 4.
05 INCHES.
THE DEPTH Can be Y, That's About 14.
19 INCHES, AND THE HEIGHT Could be Z, And that is Close to six.
08 INCHES.
Let us End BY Investigating OUR COST Functionality WHERE WE Contain the COST FUNCTION When it comes to Y AND Z.
IN 3 DIMENSIONS This might BE THE SURFACE Wherever THESE Decrease AXES Will be THE Y AND Z AXIS AND THE COST Could be Alongside THE VERTICAL AXIS.
We will SEE There is a Lower Position Right here AND THAT Happened AT OUR Vital Stage THAT WE Observed.
I HOPE YOU Identified THIS Handy.