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AN Open up Major RECTANGULAR BOX IS Getting Manufactured To carry A VOLUME OF 350 CUBIC INCHES.
THE BASE On the BOX IS Produced from Product COSTING 6 CENTS For each SQUARE INCH.
THE FRONT OF THE BOX Needs to be DECORATED And may Value twelve CENTS For every Sq. INCH.
THE REMAINDER OF The edges WILL COST 2 CENTS For each SQUARE INCH.
Discover The scale That could Lessen The price of Developing THIS BOX.
LET'S To start with DIAGRAM THE BOX AS WE SEE HERE In which The scale ARE X BY Y BY Z AND BECAUSE The amount Needs to be 350 CUBIC INCHES We now have A CONSTRAINT THAT X x Y x Z Need to Equivalent 350.
BUT Right before WE Discuss OUR Expense Functionality LETS TALK ABOUT THE SURFACE Space On the BOX.
As the Major IS OPEN, WE ONLY HAVE 5 FACES.
Let us Discover the Spot Of your 5 FACES That might MAKE UP THE Surface area Space.
Recognize The realm OF THE Entrance Experience Will be X x Z WHICH WOULD Even be Similar to THE AREA From the BACK SO THE Floor Space HAS TWO XZ Phrases.
See The correct Aspect OR The best Deal with WOULD HAVE Region Y x Z WHICH WILL BE THE Exact AS THE Still left.
Hence the Area Space Includes TWO YZ Conditions After which FINALLY The underside HAS AN AREA OF X x Y And since The very best IS Open up WE Have only One particular XY TERM IN THE Area Place AND NOW WE'LL CONVERT THE Surface area Spot TO THE COST EQUATION.
BECAUSE THE Base Price tag six CENTS PER SQUARE INCH WHERE The world OF THE BOTTOM IS X x Y Recognize HOW FOR The expense Purpose WE MULTIPLY THE XY Expression BY 6 CENTS And since THE Entrance Charges 12 CENTS For every Sq. INCH The place The world With the Entrance WOULD BE X x Z We will MULTIPLY THIS XZ TERM BY 12 CENTS IN The associated fee Functionality.
THE REMAINING SIDES Price tag 2 CENTS PER Sq. INCH SO THESE A few Parts ARE ALL MULTIPLIED BY 0.
02 OR two CENTS.
COMBINING LIKE Conditions Now we have THIS Expense FUNCTION HERE.
BUT Recognize HOW We've A few UNKNOWNS During this EQUATION SO NOW We will USE A CONSTRAINT TO FORM A price EQUATION WITH TWO VARIABLES.
IF WE Remedy OUR CONSTRAINT FOR X BY DIVIDING BOTH SIDES BY YZ WE Might make A SUBSTITUTION FOR X INTO OUR Charge Operate The place We can easily SUBSTITUTE THIS FRACTION Listed here FOR X Listed here AND Right here.
IF WE Try this, WE GET THIS EQUATION In this article AND IF WE SIMPLIFY Observe HOW THE Issue OF Z SIMPLIFIES OUT AND Listed here Variable OF Y SIMPLIFIES OUT.
SO FOR THIS FIRST Time period IF We discover THIS Product or service And after that Transfer THE Y UP WE Might have 49Y For the -1 Then FOR The final TERM IF WE Discovered THIS PRODUCT AND MOVED THE Z UP WE'D HAVE + 21Z Into the -1.
SO NOW OUR Intention IS To attenuate THIS Price Perform.
SO FOR The following Move We are going to Discover the Vital POINTS.
Essential POINTS ARE Exactly where THE Purpose Will HAVE MAX OR MIN Perform VALUES Plus they Manifest In which The main Buy OF PARTIAL DERIVATIVES ARE Both of those Equivalent TO ZERO OR Exactly where Possibly Will not EXIST.
THEN As soon as WE FIND THE Vital POINTS, We will Decide Irrespective of whether We have now A MAX Or possibly a MIN Price Utilizing OUR 2nd Get OF PARTIAL DERIVATIVES.
SO ON THIS SLIDE We are FINDING The two The 1st Purchase AND SECOND Purchase OF PARTIAL DERIVATIVES.
WE Ought to be A little bit Mindful Below However BECAUSE OUR Functionality IS A Purpose OF Y AND Z NOT X AND Y LIKE We are Utilized to.
SO FOR THE FIRST PARTIAL WITH RESPECT TO Y We might DIFFERENTIATE WITH Regard TO Y Dealing with Z AS A relentless WHICH WOULD GIVE US THIS PARTIAL Spinoff HERE.
FOR The very first PARTIAL WITH RESPECT TO Z WE WOULD DIFFERENTIATE WITH Regard TO Z AND TREAT Y AS A CONSTANT WHICH WOULD GIVE US This primary Get OF PARTIAL DERIVATIVE.
NOW USING THESE 1st ORDER OF PARTIAL DERIVATIVES WE Can discover THESE 2nd Purchase OF PARTIAL DERIVATIVES Wherever To seek out The 2nd PARTIALS WITH RESPECT TO Y We'd DIFFERENTIATE THIS PARTIAL Spinoff WITH RESPECT TO Y Yet again Offering US THIS.
THE SECOND PARTIAL WITH RESPECT TO Z We might DIFFERENTIATE THIS PARTIAL DERIVATIVE WITH RESPECT TO Z Once again Offering US THIS.
Detect HOW IT'S GIVEN Utilizing a Destructive EXPONENT As well as in Portion Kind AND THEN Lastly FOR THE Combined PARTIAL OR The 2nd Purchase OF PARTIAL WITH Regard TO Y And afterwards Z We'd DIFFERENTIATE THIS PARTIAL WITH Regard TO Z WHICH See HOW It might JUST GIVE US 0.
04.
SO NOW We will SET The primary Purchase OF PARTIAL DERIVATIVES EQUAL TO ZERO AND Remedy AS A Process OF EQUATIONS.
SO HERE ARE The very first Purchase OF PARTIALS SET Equivalent TO ZERO.
THIS IS A FAIRLY Associated SYSTEM OF EQUATIONS WHICH We will Address Making use of SUBSTITUTION.
SO I DECIDED TO Address The very first EQUATION HERE FOR Z.
SO I ADDED THIS Time period TO Either side With the EQUATION And after that DIVIDED BY 0.
04 Providing US THIS VALUE Right here FOR Z But when We discover THIS QUOTIENT AND MOVE Y For the -2 For the DENOMINATOR WE Might also WRITE Z AS THIS FRACTION Below.
Given that WE KNOW Z IS Equivalent TO THIS Portion, We could SUBSTITUTE THIS FOR Z INTO The next EQUATION Listed here.
That's WHAT WE SEE In this article BUT Discover HOW This is certainly RAISED Into the EXPONENT OF -2 SO THIS WOULD BE 1, 225 Towards the -2 DIVIDED BY Y To your -four.
SO WE Usually takes THE RECIPROCAL WHICH WOULD GIVE US Y Into the 4th DIVIDED BY 1, five hundred, 625 AND This is THE 21.
Given that WE HAVE AN EQUATION WITH Only one VARIABLE Y We wish to Address THIS FOR Y.
SO FOR The initial step, You will find there's Typical FACTOR OF Y.
SO Y = 0 WOULD Fulfill THIS EQUATION AND Could be A Vital POINT BUT WE KNOW WE'RE NOT Likely To possess a DIMENSION OF ZERO SO WE'LL JUST Dismiss THAT Worth AND Established THIS EXPRESSION In this article EQUAL TO ZERO AND Clear up WHICH IS WHAT WE SEE Right here.
SO We will ISOLATE THE Y CUBED Time period After which you can Dice ROOT Each side Of your EQUATION.
Therefore if WE Increase THIS Portion TO Each side Of your EQUATION And after that CHANGE THE Get From the EQUATION That is WHAT WE WOULD HAVE AND NOW FROM Right here TO ISOLATE Y CUBED WE Need to MULTIPLY Via the RECIPROCAL OF THIS Portion Listed here.
SO See HOW THE Remaining Facet SIMPLIFIES JUST Y CUBED Which Merchandise HERE IS Close to THIS Benefit Below.
SO NOW To resolve FOR Y We'd CUBE ROOT BOTH SIDES In the EQUATION OR Increase Each side In the EQUATION On the 1/3 Electric power AND This offers Y IS Roughly fourteen.
1918, AND NOW TO Discover the Z COORDINATE OF THE CRITICAL Position We can easily USE THIS EQUATION Below Exactly where Z = one, 225 DIVIDED BY Y SQUARED Which supplies Z IS Around 6.
0822.
We do not Need to have IT At this time BUT I WENT AHEAD And located THE CORRESPONDING X Benefit Likewise Utilizing OUR Quantity Components Remedy FOR X.
SO X Could be Roughly four.
0548.
For the reason that WE Have only A single Crucial POINT We could PROBABLY Suppose THIS Stage Will almost certainly Decrease THE COST Perform BUT TO Validate THIS WE'LL GO AHEAD AND Make use of the Essential Position AND THE SECOND Get OF PARTIAL DERIVATIVES JUST To verify.
MEANING WE'LL USE THIS Formulation Listed here FOR D AND THE VALUES OF The next Buy OF PARTIAL DERIVATIVES To ascertain WHETHER Now we have A RELATIVE MAX OR MIN AT THIS Significant Position WHEN Y IS APPROXIMATELY fourteen.
19 AND Z IS Close to six.
08.
Here's The 2nd Get OF PARTIALS THAT WE FOUND Previously.
SO We will BE SUBSTITUTING THIS Benefit FOR Y Which Price FOR Z INTO The 2nd ORDER OF PARTIALS.
WE Needs to be A bit Very careful However For the reason that REMEMBER Now we have A Purpose OF Y AND Z NOT X AND Y LIKE WE NORMALLY WOULD SO THESE X'S Could well be THESE Y'S AND THESE Y'S Might be THE Z'S.
SO The 2nd Buy OF PARTIALS WITH Regard TO Y IS Right here.
The 2nd Buy OF PARTIAL WITH RESPECT TO Z IS Below.
This is THE MIXED PARTIAL SQUARED.
Recognize The way it Will come OUT To the Optimistic Price.
Therefore if D IS Constructive AND SO IS THE SECOND PARTIAL WITH Regard TO Y Investigating OUR NOTES Right here Which means WE HAVE A RELATIVE Least AT OUR Crucial Stage And as a consequence These are generally THE DIMENSIONS THAT WOULD Limit The expense of OUR BOX.
THIS WAS THE X COORDINATE FROM THE Former SLIDE.
Here is THE Y COORDINATE AND HERE'S THE Z COORDINATE WHICH AGAIN ARE The scale OF OUR BOX.
Therefore the Entrance WIDTH Could well be X And that is Somewhere around 4.
05 INCHES.
THE DEPTH Could be Y, WHICH IS Somewhere around fourteen.
19 INCHES, AND THE HEIGHT Will be Z, WHICH IS Close to 6.
08 INCHES.
Let us End BY Investigating OUR Price tag Operate The place WE Possess the Price FUNCTION With regard to Y AND Z.
IN THREE Proportions THIS WOULD BE THE Area Where by THESE Lessen AXES Might be THE Y AND Z AXIS AND The fee Could well be Together THE VERTICAL AXIS.
WE CAN SEE THERE'S A Minimal Place HERE AND THAT OCCURRED AT OUR Vital Level THAT WE Discovered.
I HOPE YOU Observed THIS Practical.