ex-minimize-cost-to-make-open-top-box-function-of-two-variables

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AN OPEN Leading RECTANGULAR BOX IS BEING Produced To carry A VOLUME OF 350 CUBIC INCHES.

The bottom In the BOX IS Made out of Product COSTING six CENTS PER Sq. INCH.

THE FRONT In the BOX Need to be DECORATED And may Price tag twelve CENTS For every SQUARE INCH.

The rest OF The edges WILL COST 2 CENTS For every SQUARE INCH.

FIND THE DIMENSIONS THAT WILL Decrease The expense of CONSTRUCTING THIS BOX.

Let us Very first DIAGRAM THE BOX AS WE SEE HERE The place The size ARE X BY Y BY Z And since The amount Needs to be 350 CUBIC INCHES WE HAVE A CONSTRAINT THAT X x Y x Z Will have to EQUAL 350.

BUT Prior to WE TALK ABOUT OUR Expense Operate LETS Discuss THE Surface area Spot On the BOX.

BECAUSE THE Best IS Open up, WE ONLY HAVE 5 FACES.

Let us FIND THE Region With the 5 FACES That will MAKE UP THE Floor AREA.

See THE AREA On the FRONT FACE Could be X x Z Which might ALSO BE Similar to THE AREA While in the Back again And so the Area Location HAS TWO XZ Phrases.

Detect The appropriate Facet OR The ideal Encounter WOULD HAVE Region Y x Z WHICH Would be the Similar Since the LEFT.

And so the Surface area AREA Has TWO YZ Phrases After which you can Lastly THE BOTTOM HAS A place OF X x Y And since The best IS OPEN WE ONLY HAVE One particular XY Phrase IN THE SURFACE Space AND NOW We are going to Change THE Floor Place TO The associated fee EQUATION.

As the Base Value six CENTS For every Sq. INCH Wherever The region OF THE BOTTOM IS X x Y Observe HOW FOR THE COST FUNCTION WE MULTIPLY THE XY Time period BY six CENTS And since THE FRONT Charges twelve CENTS For every Sq. INCH Wherever The region On the Entrance Will be X x Z We are going to MULTIPLY THIS XZ TERM BY 12 CENTS IN The expense Functionality.

THE REMAINING SIDES COST two CENTS PER Sq. INCH SO THESE THREE Spots ARE ALL MULTIPLIED BY 0.

02 OR two CENTS.

COMBINING LIKE TERMS We've got THIS Expense Purpose HERE.

BUT Recognize HOW We now have THREE UNKNOWNS On 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 Either side BY YZ WE Will make A SUBSTITUTION FOR X INTO OUR Charge Functionality WHERE WE CAN SUBSTITUTE THIS FRACTION Right here FOR X HERE AND In this article.

IF WE Try this, WE GET THIS EQUATION In this article And when WE SIMPLIFY Recognize HOW THE Issue OF Z SIMPLIFIES OUT AND In this article Aspect OF Y SIMPLIFIES OUT.

SO FOR This primary Expression IF WE FIND THIS Item And after that Transfer THE Y UP WE WOULD HAVE 49Y On the -1 After which you can FOR THE LAST Time period IF WE FOUND THIS Product or service AND MOVED THE Z UP We would HAVE + 21Z TO THE -1.

SO NOW OUR Aim IS To attenuate THIS Expense Functionality.

SO FOR THE NEXT Action We are going to Locate the Crucial Factors.

Significant POINTS ARE Where by THE FUNCTION Will probably HAVE MAX OR MIN Operate VALUES And so they Arise WHERE The 1st ORDER OF PARTIAL DERIVATIVES ARE Equally EQUAL TO ZERO OR The place EITHER Won't EXIST.

THEN As soon as WE Discover the Vital Details, WE'LL Figure out No matter if We've A MAX OR A MIN Benefit Employing OUR 2nd Buy OF PARTIAL DERIVATIVES.

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

WE Must be A little bit Very careful Right here While Simply because OUR Perform Is actually a Perform OF Y AND Z NOT X AND Y LIKE We are Accustomed to.

SO FOR The main PARTIAL WITH Regard TO Y WE WOULD DIFFERENTIATE WITH RESPECT TO Y TREATING Z AS A continuing WHICH WOULD GIVE US THIS PARTIAL By-product Below.

FOR The initial PARTIAL WITH Regard TO Z We might DIFFERENTIATE WITH Regard TO Z AND Address Y AS A CONSTANT Which might GIVE US THIS FIRST Purchase OF PARTIAL Spinoff.

NOW Employing THESE 1st ORDER OF PARTIAL DERIVATIVES WE Can discover THESE SECOND Purchase OF PARTIAL DERIVATIVES The place To uncover THE SECOND PARTIALS WITH Regard TO Y We might DIFFERENTIATE THIS PARTIAL By-product WITH Regard TO Y Once more Providing US THIS.

The 2nd PARTIAL WITH Regard TO Z We'd DIFFERENTIATE THIS PARTIAL By-product WITH RESPECT TO Z Yet again Offering US THIS.

Recognize The way it'S GIVEN Employing a Destructive EXPONENT AND IN FRACTION Sort And after that Lastly With the Combined PARTIAL OR THE SECOND Get OF PARTIAL WITH Regard TO Y After which you can Z We might DIFFERENTIATE THIS PARTIAL WITH RESPECT TO Z WHICH Observe HOW It will JUST GIVE US 0.

04.

SO NOW WE'RE GOING TO Established The 1st Get OF PARTIAL DERIVATIVES Equivalent TO ZERO AND Clear up As being a Method OF EQUATIONS.

SO Listed below are The primary Buy OF PARTIALS SET EQUAL TO ZERO.

THIS Is a reasonably INVOLVED Technique OF EQUATIONS WHICH We are going to Remedy Employing SUBSTITUTION.

SO I Chose to Fix The primary EQUATION Listed here FOR Z.

SO I Extra THIS Phrase TO Each side Of your EQUATION And afterwards DIVIDED BY 0.

04 GIVING US THIS Benefit Listed here FOR Z But when WE FIND THIS QUOTIENT AND MOVE Y On the -two To your DENOMINATOR WE CAN ALSO WRITE Z AS THIS FRACTION In this article.

NOW THAT WE KNOW Z IS Equivalent TO THIS Portion, We could SUBSTITUTE THIS FOR Z INTO THE SECOND EQUATION Listed here.

And that is WHAT WE SEE HERE BUT Discover HOW This can be Elevated Towards the EXPONENT OF -2 SO This might BE 1, 225 TO THE -2 DIVIDED BY Y Towards the -four.

SO WE Will take THE RECIPROCAL WHICH WOULD GIVE US Y On the 4th DIVIDED BY 1, 500, 625 AND This is THE 21.

NOW THAT We've AN EQUATION WITH JUST ONE VARIABLE Y We wish to Fix THIS FOR Y.

SO FOR Step one, There's a Frequent Aspect OF Y.

SO Y = 0 WOULD SATISFY THIS EQUATION AND Might be A CRITICAL Issue BUT WE KNOW WE'RE NOT Likely TO HAVE A DIMENSION OF ZERO SO We will JUST Overlook THAT Benefit AND SET THIS EXPRESSION In this article EQUAL TO ZERO AND Fix That's WHAT WE SEE Listed here.

SO We'll ISOLATE THE Y CUBED Phrase After which CUBE ROOT Either side With the EQUATION.

SO IF WE Insert THIS FRACTION TO BOTH SIDES From the EQUATION And after that Alter the ORDER OF THE EQUATION This is certainly WHAT WE Would've AND NOW FROM Right here TO ISOLATE Y CUBED WE Must MULTIPLY BY THE RECIPROCAL Of the Portion Right here.

SO Observe HOW THE LEFT SIDE SIMPLIFIES JUST Y CUBED AND THIS Merchandise Here's APPROXIMATELY THIS VALUE Below.

SO NOW TO SOLVE FOR Y WE WOULD Dice ROOT Either side OF THE EQUATION OR RAISE Each side Of your EQUATION TO THE 1/3 Energy AND This offers Y IS Close to 14.

1918, AND NOW TO Locate the Z COORDINATE With the Essential Stage We could USE THIS EQUATION HERE The place Z = one, 225 DIVIDED BY Y SQUARED Which supplies Z IS Roughly six.

0822.

WE DON'T Need to have IT Right this moment BUT I WENT Forward And located THE CORRESPONDING X Benefit AS WELL Employing OUR Quantity Method Remedy FOR X.

SO X WOULD BE About four.

0548.

Mainly because WE ONLY HAVE A person Vital Place WE CAN Likely ASSUME THIS POINT IS GOING TO Lower The price Functionality BUT TO Confirm THIS We will Go on and USE THE Significant Place AND THE SECOND Get OF PARTIAL DERIVATIVES JUST To verify.

Which means We will USE THIS Formulation Right here FOR D And also the VALUES OF THE SECOND Get OF PARTIAL DERIVATIVES To find out Regardless of whether We now have A RELATIVE MAX OR MIN AT THIS Significant Position WHEN Y IS APPROXIMATELY fourteen.

19 AND Z IS About six.

08.

HERE ARE THE SECOND Get OF PARTIALS THAT WE FOUND Previously.

SO We will BE SUBSTITUTING THIS Price FOR Y Which VALUE FOR Z INTO The next Get OF PARTIALS.

WE Needs to be Just a little Watchful Even though Simply because Recall Now we have A FUNCTION OF Y AND Z NOT X AND Y LIKE WE Generally WOULD SO THESE X'S WOULD BE THESE Y'S AND THESE Y'S Can be THE Z'S.

SO The 2nd Buy OF PARTIALS WITH RESPECT TO Y IS Listed here.

The 2nd Purchase OF PARTIAL WITH RESPECT TO Z IS Right here.

Here is THE Blended PARTIAL SQUARED.

NOTICE HOW IT Will come OUT To your Optimistic Benefit.

SO IF D IS Favourable AND SO IS The 2nd PARTIAL WITH Regard TO Y Taking a look at OUR NOTES Listed here THAT MEANS We now have A RELATIVE MINIMUM AT OUR Significant Stage And for that reason These are typically THE DIMENSIONS That might MINIMIZE The price of OUR BOX.

THIS WAS THE X COORDINATE From your Earlier SLIDE.

HERE'S THE Y COORDINATE AND This is THE Z COORDINATE WHICH AGAIN ARE The scale OF OUR BOX.

Therefore the FRONT WIDTH Can be X That is APPROXIMATELY 4.

05 INCHES.

THE DEPTH Could well be Y, And that is Somewhere around fourteen.

19 INCHES, AND The peak Can be Z, WHICH IS Somewhere around 6.

08 INCHES.

LET'S FINISH BY Checking out OUR Charge Perform Where by WE HAVE THE Charge Operate When it comes to Y AND Z.

IN 3 Proportions THIS WOULD BE THE Surface area Exactly where THESE Reduce AXES WOULD BE THE Y AND Z AXIS AND THE COST Will be Alongside THE VERTICAL AXIS.

We are able to SEE There is a Minimal Stage HERE AND THAT OCCURRED AT OUR Vital Place THAT WE Uncovered.

I HOPE YOU Located THIS Beneficial.