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AN OPEN Best RECTANGULAR BOX IS BEING Manufactured To carry A Quantity OF 350 CUBIC INCHES.
THE BASE With the BOX IS MADE FROM MATERIAL COSTING 6 CENTS PER SQUARE INCH.
THE FRONT From the BOX MUST BE DECORATED And can Price twelve CENTS PER Sq. INCH.
THE REMAINDER OF The edges WILL Price tag 2 CENTS PER SQUARE INCH.
Locate The size That should Lower The expense of Setting up THIS BOX.
LET'S FIRST DIAGRAM THE BOX AS WE SEE Right here Where by The scale ARE X BY Y BY Z AND BECAUSE THE VOLUME MUST BE 350 CUBIC INCHES Now we have A CONSTRAINT THAT X x Y x Z Ought to EQUAL 350.
BUT In advance of WE Take a look at OUR Charge FUNCTION Allows TALK ABOUT THE Area Location From the BOX.
BECAUSE THE Best IS OPEN, WE Have only 5 FACES.
Let us Discover the Spot OF THE five FACES That will MAKE UP THE SURFACE Location.
Observe THE AREA OF THE FRONT Encounter Might be X x Z Which might Even be THE SAME AS The region Inside the BACK Therefore the SURFACE Region HAS TWO XZ TERMS.
Observe The ideal SIDE OR The best Experience WOULD HAVE Spot Y x Z WHICH Would be the Identical As being the Remaining.
Hence the Surface area AREA Consists of TWO YZ Phrases And after that Lastly THE BOTTOM HAS A location OF X x Y AND BECAUSE The highest IS Open up WE Have only A single XY Expression Within the Floor Place AND NOW We will Transform THE Floor Region TO The fee EQUATION.
Since the Base COST six CENTS For every SQUARE INCH The place The realm OF The underside IS X x Y Observe HOW FOR THE COST Perform WE MULTIPLY THE XY TERM BY six CENTS And since THE Entrance COSTS 12 CENTS For every Sq. INCH WHERE The region From the FRONT WOULD BE X x Z We are going to MULTIPLY THIS XZ Time period BY 12 CENTS IN THE COST FUNCTION.
THE REMAINING SIDES Charge 2 CENTS For each Sq. INCH SO THESE A few Places ARE ALL MULTIPLIED BY 0.
02 OR 2 CENTS.
COMBINING LIKE Phrases We've got THIS Price tag FUNCTION Below.
BUT NOTICE HOW WE HAVE A few UNKNOWNS Within this EQUATION SO NOW We will Make use of a CONSTRAINT TO Variety A price 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 Operate In which We will SUBSTITUTE THIS FRACTION Listed here FOR X In this article AND Listed here.
IF WE Do that, WE GET THIS EQUATION Right here AND IF WE SIMPLIFY See HOW THE Component OF Z SIMPLIFIES OUT AND In this article FACTOR OF Y SIMPLIFIES OUT.
SO FOR THIS FIRST Phrase IF WE FIND THIS Merchandise And afterwards MOVE THE Y UP WE Might have 49Y Towards the -one And after that FOR THE LAST Time period IF WE Discovered THIS Merchandise AND MOVED THE Z UP WE'D HAVE + 21Z TO THE -one.
SO NOW OUR Intention IS To reduce THIS Charge Operate.
SO FOR The following Move We will FIND THE Vital Factors.
Important Factors ARE The place THE Perform IS GOING TO HAVE MAX OR MIN Perform VALUES Plus they OCCUR In which The primary Get OF PARTIAL DERIVATIVES ARE Each Equivalent TO ZERO OR Exactly where EITHER Isn't going to EXIST.
THEN When WE Locate the Important Details, We are going to Figure out WHETHER We have now A MAX Or simply a MIN VALUE Utilizing OUR 2nd Get OF PARTIAL DERIVATIVES.
SO ON THIS SLIDE We are FINDING The two The very first Buy AND Next Get OF PARTIAL DERIVATIVES.
WE HAVE TO BE A little bit CAREFUL Listed here While Since OUR Purpose Can be a FUNCTION OF Y AND Z NOT X AND Y LIKE We are Utilized to.
SO FOR The initial PARTIAL WITH RESPECT TO Y We'd DIFFERENTIATE WITH RESPECT TO Y Dealing with Z AS A continuing Which might GIVE US THIS PARTIAL By-product Listed here.
FOR The initial PARTIAL WITH Regard TO Z WE WOULD DIFFERENTIATE WITH Regard TO Z AND TREAT Y AS A relentless WHICH WOULD GIVE US This primary Purchase OF PARTIAL DERIVATIVE.
NOW Applying THESE FIRST Purchase OF PARTIAL DERIVATIVES WE CAN FIND THESE 2nd Purchase OF PARTIAL DERIVATIVES Where by To locate The 2nd PARTIALS WITH RESPECT TO Y We'd DIFFERENTIATE THIS PARTIAL By-product WITH Regard TO Y AGAIN Providing US THIS.
The 2nd PARTIAL WITH RESPECT TO Z We'd DIFFERENTIATE THIS PARTIAL DERIVATIVE WITH Regard TO Z AGAIN Offering US THIS.
Discover HOW IT'S Provided USING A Adverse EXPONENT AND IN FRACTION Sort And after that Eventually FOR THE Blended PARTIAL OR The 2nd Get OF PARTIAL WITH RESPECT TO Y AND THEN Z WE WOULD DIFFERENTIATE THIS PARTIAL WITH Regard TO Z WHICH Observe HOW IT WOULD JUST GIVE US 0.
04.
SO NOW We will Established THE FIRST Get OF PARTIAL DERIVATIVES EQUAL TO ZERO AND Resolve Being a Program OF EQUATIONS.
SO Here i will discuss The 1st Get OF PARTIALS Established EQUAL TO ZERO.
THIS IS A FAIRLY Associated SYSTEM OF EQUATIONS WHICH We are going to Resolve Working with SUBSTITUTION.
SO I Made a decision to SOLVE The very first EQUATION HERE FOR Z.
SO I Extra THIS Phrase TO Either side From the EQUATION After which DIVIDED BY 0.
04 Offering US THIS Benefit HERE FOR Z BUT IF We discover THIS QUOTIENT AND MOVE Y Into the -2 Towards the DENOMINATOR WE Could also Generate Z AS THIS FRACTION Listed here.
Given that We all know Z IS Equivalent TO THIS Portion, We are able to SUBSTITUTE THIS FOR Z INTO The next EQUATION Below.
WHICH IS WHAT WE SEE In this article BUT NOTICE HOW This is often RAISED Into the EXPONENT OF -2 SO This might BE one, 225 On the -two DIVIDED BY Y For the -four.
SO WE CAN TAKE THE RECIPROCAL Which might GIVE US Y Into the 4th DIVIDED BY one, five hundred, 625 AND This is THE 21.
NOW THAT We've AN EQUATION WITH Only one VARIABLE Y We wish to Remedy THIS FOR Y.
SO FOR Step one, There's a Typical Aspect OF Y.
SO Y = 0 WOULD SATISFY THIS EQUATION AND Can be A Essential Issue BUT WE KNOW We are NOT Heading To possess a DIMENSION OF ZERO SO We are going to JUST Disregard THAT VALUE AND SET THIS EXPRESSION In this article Equivalent TO ZERO AND Remedy That's WHAT WE SEE In this article.
SO We will ISOLATE THE Y CUBED Phrase After which Dice ROOT Either side Of your EQUATION.
SO IF WE Incorporate THIS FRACTION TO Each side On the EQUATION AND THEN Alter the Purchase With the EQUATION This is certainly WHAT WE Would've AND NOW FROM HERE TO ISOLATE Y CUBED WE HAVE TO MULTIPLY Via the RECIPROCAL Of the FRACTION Right here.
SO Detect HOW THE LEFT SIDE SIMPLIFIES JUST Y CUBED AND THIS Merchandise Here's Roughly THIS Price HERE.
SO NOW TO SOLVE FOR Y We'd Dice ROOT Either side With the EQUATION OR RAISE Each side With the EQUATION Towards the 1/three Energy AND This offers Y IS About 14.
1918, AND NOW TO Locate the Z COORDINATE From the Crucial POINT We will USE THIS EQUATION Right here Wherever Z = 1, 225 DIVIDED BY Y SQUARED Which provides Z IS APPROXIMATELY six.
0822.
We do not Have to have IT Today BUT I WENT In advance And located THE CORRESPONDING X VALUE Likewise Making use of OUR Quantity Formulation Fix FOR X.
SO X Might be APPROXIMATELY four.
0548.
Due to the fact WE ONLY HAVE One particular CRITICAL Stage WE CAN In all probability Believe THIS Issue Will Decrease The fee FUNCTION BUT TO VERIFY THIS We are going to Go on and Utilize the Essential Issue AND The 2nd Buy OF PARTIAL DERIVATIVES JUST To make certain.
MEANING We will USE THIS Method Below FOR D Along with the VALUES OF The next Buy OF PARTIAL DERIVATIVES To ascertain Whether or not We now have A RELATIVE MAX OR MIN AT THIS Crucial Position WHEN Y IS Close to fourteen.
19 AND Z IS APPROXIMATELY 6.
08.
HERE ARE The 2nd Get OF PARTIALS THAT WE Uncovered EARLIER.
SO WE'LL BE SUBSTITUTING THIS Worth FOR Y AND THIS Benefit FOR Z INTO The next Buy OF PARTIALS.
WE Ought to be A bit CAREFUL Nevertheless BECAUSE Try to remember We've A Operate OF Y AND Z NOT X AND Y LIKE WE NORMALLY WOULD SO THESE X'S Can be THESE Y'S AND THESE Y'S Will be THE Z'S.
SO The next Purchase OF PARTIALS WITH RESPECT TO Y IS HERE.
THE SECOND Get OF PARTIAL WITH RESPECT TO Z IS Below.
HERE'S THE Blended PARTIAL SQUARED.
Recognize The way it Arrives OUT To the POSITIVE Price.
SO IF D IS Beneficial AND SO IS THE SECOND PARTIAL WITH RESPECT TO Y Thinking about OUR NOTES HERE Meaning We've got A RELATIVE Bare minimum AT OUR Crucial Position And thus These are generally The scale That might MINIMIZE The expense of OUR BOX.
THIS WAS THE X COORDINATE Through the Earlier SLIDE.
Here is THE Y COORDINATE AND This is THE Z COORDINATE WHICH All over again ARE THE DIMENSIONS OF OUR BOX.
Therefore the FRONT WIDTH Could be X That is Close to four.
05 INCHES.
THE DEPTH Can be Y, That's Close to fourteen.
19 INCHES, AND THE HEIGHT Could well be Z, That's Somewhere around 6.
08 INCHES.
Let us End BY Thinking about OUR Value Operate Where by WE Possess the Price tag FUNCTION Regarding Y AND Z.
IN 3 DIMENSIONS This is able to BE THE Area The place THESE Decrease AXES Might be THE Y AND Z AXIS AND The fee Might be Together THE VERTICAL AXIS.
We can easily SEE THERE'S A Very low Place Listed here Which Transpired AT OUR Essential Level THAT WE Identified.
I HOPE YOU Observed THIS Handy.