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AN OPEN Prime RECTANGULAR BOX IS Getting CONSTRUCTED To carry A VOLUME OF 350 CUBIC INCHES.
The bottom With the BOX IS Produced from Substance COSTING 6 CENTS PER SQUARE INCH.
THE FRONT With the BOX Should be DECORATED AND WILL Value twelve CENTS For each Sq. INCH.
The rest OF The edges WILL Price two CENTS PER SQUARE INCH.
FIND The scale That can MINIMIZE The expense of Setting up THIS BOX.
Let us Very first DIAGRAM THE BOX AS WE SEE In this article Exactly where The scale ARE X BY Y BY Z AND BECAUSE The quantity Need to be 350 CUBIC INCHES We have now A CONSTRAINT THAT X x Y x Z Have to Equivalent 350.
BUT Just before WE Look at OUR COST Perform Allows Take a look at THE Area Region With the BOX.
As the TOP IS Open up, WE Have only 5 FACES.
LET'S Locate the Place With the 5 FACES That may MAKE UP THE Area Location.
Observe The realm Of your Entrance Facial area Might be X x Z WHICH WOULD ALSO BE Similar to THE AREA During the Again And so the SURFACE AREA HAS TWO XZ TERMS.
NOTICE The best Aspect OR The appropriate Facial area Might have Space Y x Z WHICH Would be the Exact same AS THE Still left.
Therefore the SURFACE Space Includes TWO YZ Conditions And after that FINALLY THE BOTTOM HAS A region OF X x Y AND BECAUSE The very best IS Open up WE ONLY HAVE A person XY Time period Inside the Area Place AND NOW We are going to Change THE Area Spot TO The associated fee EQUATION.
As the Base Price 6 CENTS For each SQUARE INCH In which THE AREA OF THE BOTTOM IS X x Y See HOW FOR The price FUNCTION WE MULTIPLY THE XY Time period BY six CENTS And since THE FRONT Charges twelve CENTS PER SQUARE INCH Wherever The region With the FRONT Might be X x Z WE'LL MULTIPLY THIS XZ TERM BY 12 CENTS IN The expense Perform.
THE REMAINING SIDES Charge two CENTS For every SQUARE INCH SO THESE THREE Parts ARE ALL MULTIPLIED BY 0.
02 OR 2 CENTS.
COMBINING LIKE TERMS We now have THIS COST Perform Right here.
BUT Detect HOW Now we have THREE UNKNOWNS With this EQUATION SO NOW We are going to Utilize a CONSTRAINT TO FORM A price EQUATION WITH TWO VARIABLES.
IF WE Remedy OUR CONSTRAINT FOR X BY DIVIDING Each side BY YZ WE May make A SUBSTITUTION FOR X INTO OUR Price Functionality Where by WE CAN SUBSTITUTE THIS FRACTION Right here FOR X Right here AND HERE.
IF WE Do that, WE GET THIS EQUATION Right here AND IF WE SIMPLIFY NOTICE HOW THE Issue OF Z SIMPLIFIES OUT AND Right here Issue OF Y SIMPLIFIES OUT.
SO FOR THIS FIRST Time period IF We discover THIS Solution And after that MOVE THE Y UP WE Would've 49Y On the -one And afterwards FOR THE LAST TERM IF WE Identified THIS Item AND MOVED THE Z UP We might HAVE + 21Z Towards the -one.
SO NOW OUR Objective IS To attenuate THIS Expense Operate.
SO FOR The following Action We will Locate the Crucial Details.
Significant Details ARE The place THE Functionality Will HAVE MAX OR MIN Operate VALUES AND THEY Arise Exactly where The main ORDER OF PARTIAL DERIVATIVES ARE The two Equivalent TO ZERO OR In which Both DOES NOT EXIST.
THEN As soon as WE Locate the Crucial Details, We are going to Establish WHETHER We now have A MAX Or simply a MIN Price Working with OUR SECOND ORDER OF PARTIAL DERIVATIVES.
SO ON THIS SLIDE WE'RE Locating The two The main ORDER AND Next Buy OF PARTIAL DERIVATIVES.
WE Must be A little bit Watchful Listed here THOUGH Due to the fact OUR Purpose Is really a Purpose OF Y AND Z NOT X AND Y LIKE We are USED TO.
SO FOR THE FIRST PARTIAL WITH RESPECT TO Y WE WOULD DIFFERENTIATE WITH RESPECT TO Y Dealing with Z AS A relentless WHICH WOULD GIVE US THIS PARTIAL DERIVATIVE Right here.
FOR The 1st PARTIAL WITH RESPECT TO Z We might DIFFERENTIATE WITH Regard TO Z AND Take care of Y AS A CONSTANT WHICH WOULD GIVE US This primary Buy OF PARTIAL By-product.
NOW Making use of THESE Initially Purchase OF PARTIAL DERIVATIVES WE Can discover THESE 2nd Purchase OF PARTIAL DERIVATIVES In which To search out The 2nd PARTIALS WITH RESPECT TO Y We might DIFFERENTIATE THIS PARTIAL Spinoff WITH Regard TO Y AGAIN Offering US THIS.
The 2nd PARTIAL WITH Regard TO Z We might DIFFERENTIATE THIS PARTIAL Spinoff WITH RESPECT TO Z Yet again Offering US THIS.
See The way it'S Specified Utilizing a Destructive EXPONENT As well as in Portion Variety AND THEN Eventually To the MIXED PARTIAL OR The next Purchase OF PARTIAL WITH Regard TO Y After which Z We'd DIFFERENTIATE THIS PARTIAL WITH RESPECT TO Z WHICH See HOW It could JUST GIVE US 0.
04.
SO NOW We'll Established The initial Get OF PARTIAL DERIVATIVES EQUAL TO ZERO AND Address As being a Procedure OF EQUATIONS.
SO Listed here are The primary ORDER OF PARTIALS Established Equivalent TO ZERO.
THIS Is a reasonably Associated Process OF EQUATIONS WHICH We are going to Resolve Making use of SUBSTITUTION.
SO I Chose to Address The 1st EQUATION HERE FOR Z.
SO I Additional THIS Phrase TO Either side On the EQUATION AND THEN DIVIDED BY 0.
04 GIVING US THIS Price Right here FOR Z However, if WE FIND THIS QUOTIENT AND Shift Y To your -2 Towards the DENOMINATOR WE CAN ALSO Create Z AS THIS FRACTION Below.
Given that We all know Z IS EQUAL TO THIS Portion, We can easily SUBSTITUTE THIS FOR Z INTO THE SECOND EQUATION Listed here.
And that is WHAT WE SEE Below BUT Observe HOW This can be Lifted On the EXPONENT OF -two SO This is able to BE one, 225 TO THE -two DIVIDED BY Y For the -4.
SO WE CAN TAKE THE RECIPROCAL Which might GIVE US Y To your 4th DIVIDED BY 1, 500, 625 AND Here is THE 21.
Given that WE HAVE AN EQUATION WITH Only one VARIABLE Y We wish to Clear up THIS FOR Y.
SO FOR The initial step, There's a Prevalent Aspect OF Y.
SO Y = 0 WOULD Fulfill THIS EQUATION AND Can be A Important Level BUT WE KNOW WE'RE NOT Heading TO HAVE A DIMENSION OF ZERO SO We are going to JUST Overlook THAT VALUE AND SET THIS EXPRESSION In this article Equivalent TO ZERO AND Resolve That is WHAT WE SEE Listed here.
SO We'll ISOLATE THE Y CUBED Time period Then CUBE ROOT BOTH SIDES OF THE EQUATION.
SO IF WE Include THIS FRACTION TO Each side Of your EQUATION And after that Alter the Purchase In the EQUATION This is certainly WHAT WE Would've AND NOW FROM In this article TO ISOLATE Y CUBED WE Need to MULTIPLY Through the RECIPROCAL OF THIS FRACTION HERE.
SO Observe HOW THE Still left Facet SIMPLIFIES JUST Y CUBED Which Item HERE IS APPROXIMATELY THIS VALUE Listed here.
SO NOW To resolve FOR Y WE WOULD CUBE ROOT BOTH SIDES OF THE EQUATION OR RAISE Either side On the EQUATION For the one/3 Electric power AND This provides Y IS Roughly 14.
1918, AND NOW TO FIND THE Z COORDINATE With the Important POINT We can easily USE THIS EQUATION HERE Where by Z = 1, 225 DIVIDED BY Y SQUARED Which provides Z IS Close to 6.
0822.
We do not Want IT At the moment BUT I WENT In advance AND FOUND THE CORRESPONDING X VALUE AS WELL Utilizing OUR Quantity Method Resolve FOR X.
SO X WOULD BE About four.
0548.
BECAUSE WE Have only A person Crucial Issue WE CAN PROBABLY ASSUME THIS Position Will Lessen The price Functionality BUT TO VERIFY THIS We will Go on and USE THE Essential Level AND THE SECOND Buy OF PARTIAL DERIVATIVES JUST To be sure.
MEANING We will USE THIS System In this article FOR D Plus the VALUES OF THE SECOND Purchase OF PARTIAL DERIVATIVES To ascertain WHETHER We now have A RELATIVE MAX OR MIN AT THIS CRITICAL Position WHEN Y IS Close to 14.
19 AND Z IS Close to 6.
08.
Allow me to share The 2nd Get OF PARTIALS THAT WE Located Before.
SO WE'LL BE SUBSTITUTING THIS Worth FOR Y Which Worth FOR Z INTO The 2nd Get OF PARTIALS.
WE Needs to be Just a little Watchful While Due to the fact REMEMBER We've got A Operate OF Y AND Z NOT X AND Y LIKE WE Usually WOULD SO THESE X'S Can be THESE Y'S AND THESE Y'S WOULD BE THE Z'S.
SO The next Get OF PARTIALS WITH Regard TO Y IS In this article.
THE SECOND ORDER OF PARTIAL WITH RESPECT TO Z IS Right here.
HERE'S THE MIXED PARTIAL SQUARED.
Observe The way it Will come OUT Into a POSITIVE VALUE.
Therefore if D IS Favourable AND SO IS THE SECOND PARTIAL WITH Regard TO Y Checking out OUR NOTES In this article Which means WE HAVE A RELATIVE MINIMUM AT OUR Vital Place And so These are generally The scale That may Lessen The price of OUR BOX.
THIS WAS THE X COORDINATE Through the Prior SLIDE.
HERE'S THE Y COORDINATE AND This is THE Z COORDINATE WHICH All over again ARE The size OF OUR BOX.
Therefore the Entrance WIDTH Could well be X That's About four.
05 INCHES.
THE DEPTH Can be Y, And that is APPROXIMATELY 14.
19 INCHES, AND THE HEIGHT Might be Z, That's APPROXIMATELY six.
08 INCHES.
Let us End BY Considering OUR Charge Purpose Wherever WE Possess the Charge Functionality With regard to Y AND Z.
IN THREE Proportions This might BE THE Surface area Exactly where THESE Lessen AXES Could be THE Y AND Z AXIS AND The associated fee WOULD BE Together THE VERTICAL AXIS.
WE CAN SEE There is a Reduced Issue Right here Which Happened AT OUR Essential POINT THAT WE Located.
I HOPE YOU Observed THIS Beneficial.