Ex: Minimize Cost to Make Open Top Box – Function of Two Variables

iptv billiga Gratis 2. Upplev den bästa ... IPTVBilliga IPTV-servern är stabil, jag gillar din kanallista. ... Jag köper en IPTV IPTVBilliga prenumeration för en månad.

AN Open up Best RECTANGULAR BOX IS BEING Created To carry A Quantity OF 350 CUBIC INCHES.

The bottom From the BOX IS Created from Product COSTING 6 CENTS For each Sq. INCH.

THE Entrance From the BOX Need to be DECORATED And may COST twelve CENTS For every SQUARE INCH.

The rest OF THE SIDES WILL Value two CENTS For each Sq. INCH.

FIND The scale Which will MINIMIZE The expense of Setting up THIS BOX.

LET'S Initially DIAGRAM THE BOX AS WE SEE HERE In which The size ARE X BY Y BY Z And since The quantity Need to be 350 CUBIC INCHES We've A CONSTRAINT THAT X x Y x Z Have to Equivalent 350.

BUT BEFORE WE Mention OUR Price tag Purpose Allows Speak about THE Surface area AREA In the BOX.

As the Leading IS Open up, WE ONLY HAVE five FACES.

Let us FIND THE Region On the five FACES THAT WOULD MAKE UP THE Area Location.

See THE AREA Of your FRONT Encounter WOULD BE X x Z Which might ALSO BE THE SAME AS THE AREA Inside the Again Therefore the Area Spot HAS TWO XZ Conditions.

Discover The ideal Facet OR THE RIGHT Encounter Would've Region Y x Z WHICH Would be the SAME Given that the Remaining.

Hence the Floor AREA Includes TWO YZ Phrases And afterwards Last but not least The underside HAS A region OF X x Y And since The highest IS OPEN WE ONLY HAVE One particular XY Phrase From the SURFACE Spot AND NOW We will CONVERT THE Area Location TO The price EQUATION.

BECAUSE THE Base Price six CENTS PER SQUARE INCH Where by THE AREA OF THE BOTTOM IS X x Y Recognize HOW FOR The price Functionality WE MULTIPLY THE XY Phrase BY six CENTS And since THE Entrance Prices twelve CENTS PER Sq. INCH Exactly where The world With the Entrance Can be X x Z We are going to MULTIPLY THIS XZ Time period BY 12 CENTS IN The price Operate.

THE REMAINING SIDES Value two CENTS PER SQUARE INCH SO THESE A few Regions ARE ALL MULTIPLIED BY 0.

02 OR 2 CENTS.

COMBINING LIKE Conditions We've THIS Charge Functionality Below.

BUT Observe HOW We have now A few UNKNOWNS In this particular EQUATION SO NOW WE'LL Utilize a CONSTRAINT TO Kind A value EQUATION WITH TWO VARIABLES.

IF WE Fix OUR CONSTRAINT FOR X BY DIVIDING BOTH SIDES BY YZ WE May make A SUBSTITUTION FOR X INTO OUR Value FUNCTION Wherever WE CAN SUBSTITUTE THIS Portion HERE FOR X Listed here AND HERE.

IF WE Make this happen, WE GET THIS EQUATION Below And when WE SIMPLIFY Detect HOW THE Issue OF Z SIMPLIFIES OUT AND In this article FACTOR OF Y SIMPLIFIES OUT.

SO FOR THIS FIRST TERM IF We discover THIS Products Then Go THE Y UP WE WOULD HAVE 49Y For the -1 AND THEN FOR The final Time period IF WE Observed THIS PRODUCT AND MOVED THE Z UP We would HAVE + 21Z To your -1.

SO NOW OUR Aim IS To attenuate THIS Value FUNCTION.

SO FOR THE NEXT Phase We will Locate the Important Factors.

Important Details ARE WHERE THE Functionality Will HAVE MAX OR MIN Purpose VALUES They usually Manifest The place The 1st Buy OF PARTIAL DERIVATIVES ARE Equally EQUAL TO ZERO OR The place Possibly Isn't going to EXIST.

THEN ONCE WE Locate the Crucial Details, WE'LL Establish WHETHER We've got A MAX Or even a MIN Price Utilizing OUR Next Get OF PARTIAL DERIVATIVES.

SO ON THIS SLIDE We are Discovering BOTH The main Get AND SECOND Buy OF PARTIAL DERIVATIVES.

WE HAVE TO BE A LITTLE Mindful Right here While BECAUSE OUR Functionality Is usually a Operate OF Y AND Z NOT X AND Y LIKE We are Accustomed to.

SO FOR The main 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 Right here.

FOR The 1st PARTIAL WITH Regard TO Z We'd DIFFERENTIATE WITH Regard TO Z AND Handle Y AS A relentless WHICH WOULD GIVE US This primary Purchase OF PARTIAL By-product.

NOW Working with THESE Initial Purchase OF PARTIAL DERIVATIVES WE CAN FIND THESE 2nd Get OF PARTIAL DERIVATIVES The place To uncover THE SECOND PARTIALS WITH Regard TO Y We'd DIFFERENTIATE THIS PARTIAL DERIVATIVE WITH Regard TO Y Once again GIVING US THIS.

THE SECOND PARTIAL WITH RESPECT TO Z WE WOULD DIFFERENTIATE THIS PARTIAL DERIVATIVE WITH Regard TO Z Once again Supplying US THIS.

Recognize The way it'S Supplied Utilizing a Adverse EXPONENT As well as in Portion Sort AND THEN FINALLY To the MIXED PARTIAL OR THE SECOND Purchase OF PARTIAL WITH Regard TO Y And after that Z We might DIFFERENTIATE THIS PARTIAL WITH RESPECT TO Z WHICH See HOW It will JUST GIVE US 0.

04.

SO NOW We'll Established The 1st Buy OF PARTIAL DERIVATIVES Equivalent TO ZERO AND Address Being a Method OF EQUATIONS.

SO Allow me to share The 1st Buy OF PARTIALS Established Equivalent TO ZERO.

THIS Is a reasonably Associated Technique OF EQUATIONS WHICH We will Resolve Making use of SUBSTITUTION.

SO I Made a decision to Resolve The very first EQUATION In this article FOR Z.

SO I Extra THIS Time period TO Either side On the EQUATION And afterwards DIVIDED BY 0.

04 Supplying US THIS VALUE Right here FOR Z BUT IF We discover THIS QUOTIENT AND Shift Y Into the -two TO THE DENOMINATOR WE CAN ALSO Produce Z AS THIS Portion HERE.

Since WE KNOW Z IS EQUAL TO THIS Portion, We could SUBSTITUTE THIS FOR Z INTO THE SECOND EQUATION HERE.

Which can be WHAT WE SEE In this article BUT Detect HOW This is certainly Lifted TO THE EXPONENT OF -2 SO This may BE one, 225 Towards the -two DIVIDED BY Y To your -4.

SO WE Normally takes THE RECIPROCAL WHICH WOULD GIVE US Y Towards the 4th DIVIDED BY one, five hundred, 625 AND This is THE 21.

NOW THAT WE HAVE AN EQUATION WITH Only one VARIABLE Y WE WANT TO Remedy THIS FOR Y.

SO FOR Step one, You will find there's Popular Component OF Y.

SO Y = 0 WOULD Fulfill THIS EQUATION AND Can be A Significant POINT BUT We all know WE'RE NOT GOING To possess a DIMENSION OF ZERO SO We will JUST Disregard THAT VALUE AND Established THIS EXPRESSION Right here EQUAL TO ZERO AND Resolve Which happens to be WHAT WE SEE Right here.

SO We'll ISOLATE THE Y CUBED TERM And after that CUBE ROOT Either side With the EQUATION.

Therefore if WE Insert THIS FRACTION TO BOTH SIDES On the EQUATION Then CHANGE THE Purchase OF THE EQUATION This really is WHAT WE Might have AND NOW FROM In this article TO ISOLATE Y CUBED WE HAVE TO MULTIPLY From the RECIPROCAL Of the FRACTION In this article.

SO Observe HOW THE Still left Facet SIMPLIFIES JUST Y CUBED AND THIS Solution HERE IS Somewhere around THIS Benefit Below.

SO NOW To resolve FOR Y WE WOULD CUBE ROOT Each side With the EQUATION OR Elevate Either side With the EQUATION TO THE 1/3 Energy AND This offers Y IS Roughly fourteen.

1918, AND NOW TO FIND THE Z COORDINATE OF THE Vital Place We are able to USE THIS EQUATION Right here Wherever Z = 1, 225 DIVIDED BY Y SQUARED Which supplies Z IS Roughly six.

0822.

WE DON'T NEED IT Today BUT I WENT Forward AND FOUND THE CORRESPONDING X Worth Also Applying OUR VOLUME Formulation Address FOR X.

SO X Will be Around four.

0548.

For the reason that WE ONLY HAVE 1 Vital Place We could In all probability Suppose THIS POINT IS GOING TO Decrease The expense Functionality BUT TO Confirm THIS WE'LL Go on and Make use of the Essential Level AND THE SECOND Purchase OF PARTIAL DERIVATIVES JUST To verify.

MEANING We will USE THIS Components HERE FOR D AND THE VALUES OF The 2nd ORDER OF PARTIAL DERIVATIVES To find out WHETHER We now have A RELATIVE MAX OR MIN AT THIS CRITICAL Place WHEN Y IS About 14.

19 AND Z IS About 6.

08.

HERE ARE The next Get OF PARTIALS THAT WE Uncovered EARLIER.

SO We are going to BE SUBSTITUTING THIS VALUE FOR Y Which Worth FOR Z INTO The 2nd Buy OF PARTIALS.

WE Needs to be Somewhat Mindful THOUGH Since Don't forget Now we have A FUNCTION 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 Can be THE Z'S.

SO The next Get OF PARTIALS WITH RESPECT TO Y IS Right here.

The 2nd Get OF PARTIAL WITH Regard TO Z IS HERE.

This is THE MIXED PARTIAL SQUARED.

Discover The way it Arrives OUT To your Good Worth.

Therefore if D IS Optimistic AND SO IS The next PARTIAL WITH RESPECT TO Y Checking out OUR NOTES Right here THAT MEANS We now have A RELATIVE Minimum amount AT OUR Important Level And for that reason They are The scale That may Limit The price of OUR BOX.

THIS WAS THE X COORDINATE From your PREVIOUS SLIDE.

This is THE Y COORDINATE AND Here is THE Z COORDINATE WHICH Yet again ARE The size OF OUR BOX.

SO THE Entrance WIDTH Could be X And that is Roughly 4.

05 INCHES.

THE DEPTH WOULD BE Y, And that is Around 14.

19 INCHES, AND THE HEIGHT WOULD BE Z, And that is Around 6.

08 INCHES.

LET'S FINISH BY Checking out OUR Charge Operate Exactly where WE Hold the Price Functionality With regards to Y AND Z.

IN THREE DIMENSIONS THIS WOULD BE THE SURFACE Wherever THESE Reduce AXES Could well be THE Y AND Z AXIS AND The price Can be ALONG THE VERTICAL AXIS.

We could SEE THERE'S A Lower Issue Listed here Which Happened AT OUR Significant Level THAT WE FOUND.

I HOPE YOU Located THIS Useful.