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AN OPEN Major RECTANGULAR BOX IS Becoming Manufactured To carry A VOLUME OF 350 CUBIC INCHES.
The bottom On the BOX IS Constructed from MATERIAL COSTING six CENTS PER Sq. INCH.
THE FRONT With the BOX Needs to be DECORATED AND WILL Charge 12 CENTS For each Sq. INCH.
THE REMAINDER OF The edges WILL Price tag two CENTS For every SQUARE INCH.
Uncover The size THAT WILL MINIMIZE THE COST OF Setting up THIS BOX.
Let us 1st DIAGRAM THE BOX AS WE SEE Below In which THE DIMENSIONS ARE X BY Y BY Z AND BECAUSE THE VOLUME Need to be 350 CUBIC INCHES We've A CONSTRAINT THAT X x Y x Z Have to Equivalent 350.
BUT In advance of WE TALK ABOUT OUR Charge FUNCTION LETS Speak about THE Area AREA Of your BOX.
Since the Major IS OPEN, WE Have only 5 FACES.
LET'S FIND THE Place Of your five FACES That could MAKE UP THE SURFACE Place.
Observe The region In the FRONT Experience Could be X x Z WHICH WOULD ALSO BE Similar to The realm From the BACK SO THE Area AREA HAS TWO XZ TERMS.
Recognize The proper SIDE OR The ideal Encounter Would've Space Y x Z WHICH Would be the Similar AS THE Remaining.
Therefore the Floor Place Includes TWO YZ Conditions Then At last THE BOTTOM HAS A region OF X x Y And since THE TOP IS Open up WE ONLY HAVE Just one XY Expression During the Surface area Region AND NOW We are going to CONVERT THE SURFACE Space TO The fee EQUATION.
Since the Base Price 6 CENTS For each Sq. INCH WHERE The region OF THE BOTTOM IS X x Y Observe HOW FOR The price Functionality WE MULTIPLY THE XY Expression BY six CENTS AND BECAUSE THE Entrance Charges 12 CENTS For each SQUARE INCH The place THE AREA In the Entrance Could be X x Z We will MULTIPLY THIS XZ TERM BY 12 CENTS IN The price Functionality.
THE REMAINING SIDES Price tag two CENTS For each SQUARE INCH SO THESE THREE Spots ARE ALL MULTIPLIED BY 0.
02 OR two CENTS.
COMBINING LIKE Phrases We've got THIS Charge Functionality In this article.
BUT See HOW We've got THREE UNKNOWNS During this EQUATION SO NOW We are going to USE A CONSTRAINT TO Kind A value 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 Value Functionality Exactly where WE CAN SUBSTITUTE THIS FRACTION In this article FOR X HERE AND Listed here.
IF WE Try this, WE GET THIS EQUATION Below AND IF WE SIMPLIFY Observe HOW THE Issue OF Z SIMPLIFIES OUT AND Right here Issue OF Y SIMPLIFIES OUT.
SO FOR This primary TERM IF WE FIND THIS PRODUCT After which you can Shift THE Y UP WE Might have 49Y To your -1 AND THEN FOR The final Phrase IF WE Identified THIS PRODUCT AND MOVED THE Z UP We might HAVE + 21Z Into the -one.
SO NOW OUR Intention IS TO MINIMIZE THIS Price tag FUNCTION.
SO FOR Another Action WE'LL Discover the Crucial POINTS.
Crucial POINTS ARE In which THE Purpose Will probably HAVE MAX OR MIN Operate VALUES And so they Take place In which THE FIRST ORDER OF PARTIAL DERIVATIVES ARE BOTH Equivalent TO ZERO OR Exactly where EITHER Would not EXIST.
THEN ONCE WE Locate the Vital Factors, We will Ascertain Irrespective of whether We have now A MAX Or maybe a MIN Worth Making use of OUR 2nd Buy OF PARTIAL DERIVATIVES.
SO ON THIS SLIDE WE'RE Discovering The two The primary Buy AND SECOND Buy OF PARTIAL DERIVATIVES.
WE Ought to be Just a little CAREFUL In this article THOUGH Mainly because OUR Operate Is usually a Operate OF Y AND Z NOT X AND Y LIKE We are USED TO.
SO FOR The 1st PARTIAL WITH Regard TO Y We might DIFFERENTIATE WITH Regard TO Y TREATING Z AS A CONSTANT Which might GIVE US THIS PARTIAL By-product Below.
FOR The very first PARTIAL WITH Regard TO Z WE WOULD DIFFERENTIATE WITH RESPECT TO Z AND Deal with Y AS A relentless Which might GIVE US THIS FIRST Buy OF PARTIAL By-product.
NOW Making use of THESE Initial ORDER OF PARTIAL DERIVATIVES WE Can discover THESE SECOND Buy OF PARTIAL DERIVATIVES Wherever To seek out The next PARTIALS WITH Regard TO Y WE WOULD DIFFERENTIATE THIS PARTIAL Spinoff WITH RESPECT TO Y Yet again GIVING US THIS.
The next PARTIAL WITH Regard TO Z We might DIFFERENTIATE THIS PARTIAL Spinoff WITH Regard TO Z Once more GIVING US THIS.
NOTICE HOW IT'S Offered USING A Damaging EXPONENT As well as in Portion Type AND THEN Last but not least FOR THE Blended PARTIAL OR The next Get OF PARTIAL WITH Regard TO Y And after that Z We might DIFFERENTIATE THIS PARTIAL WITH Regard TO Z WHICH Observe HOW It will JUST GIVE US 0.
04.
SO NOW WE'RE GOING TO Established The initial Purchase OF PARTIAL DERIVATIVES Equivalent TO ZERO AND Resolve AS A Technique OF EQUATIONS.
SO HERE ARE The main Purchase OF PARTIALS Established EQUAL TO ZERO.
THIS IS A FAIRLY INVOLVED Technique OF EQUATIONS WHICH We will Address Making use of SUBSTITUTION.
SO I DECIDED TO Address THE FIRST EQUATION Listed here FOR Z.
SO I Extra THIS Time period TO BOTH SIDES OF THE EQUATION And after that DIVIDED BY 0.
04 GIVING US THIS Price In this article FOR Z However, if WE FIND THIS QUOTIENT AND Go Y Into the -2 TO THE DENOMINATOR WE CAN ALSO Publish Z AS THIS FRACTION Below.
NOW THAT WE KNOW Z IS EQUAL TO THIS FRACTION, We are able to SUBSTITUTE THIS FOR Z INTO The next EQUATION Right here.
And that is WHAT WE SEE In this article BUT Detect HOW THIS IS Elevated On the EXPONENT OF -two SO This is able to BE 1, 225 Into the -2 DIVIDED BY Y For the -4.
SO WE Usually takes THE RECIPROCAL Which might GIVE US Y On the 4th DIVIDED BY 1, five hundred, 625 AND HERE'S THE 21.
Given that We've got AN EQUATION WITH JUST ONE VARIABLE Y We wish to Remedy THIS FOR Y.
SO FOR THE FIRST STEP, There's a Typical Component OF Y.
SO Y = 0 WOULD Fulfill THIS EQUATION AND Could be A Vital Position BUT We all know We are NOT Heading TO HAVE A DIMENSION OF ZERO SO We are going to JUST Overlook THAT VALUE AND Established THIS EXPRESSION Below Equivalent TO ZERO AND Address WHICH IS WHAT WE SEE Listed here.
SO WE'RE GOING TO ISOLATE THE Y CUBED Time period AND THEN Dice ROOT Each side From the EQUATION.
Therefore if WE ADD THIS Portion TO Either side With the EQUATION And after that CHANGE THE Get From the EQUATION This really is WHAT WE WOULD HAVE AND NOW FROM HERE TO ISOLATE Y CUBED WE Really have to MULTIPLY Through the RECIPROCAL Of the Portion Right here.
SO Discover HOW THE Still left Facet SIMPLIFIES JUST Y CUBED AND THIS Solution HERE IS APPROXIMATELY THIS Worth HERE.
SO NOW To resolve FOR Y We might CUBE ROOT BOTH SIDES From the EQUATION OR RAISE Each side OF THE EQUATION To your 1/three Ability AND This offers Y IS Close to 14.
1918, AND NOW TO Locate the Z COORDINATE Of your Important Place We can easily USE THIS EQUATION In this article The place Z = one, 225 DIVIDED BY Y SQUARED Which provides Z IS Somewhere around six.
0822.
WE DON'T Need to have IT At this moment BUT I WENT AHEAD And located THE CORRESPONDING X Benefit Too Employing OUR Quantity System Fix FOR X.
SO X Could well be APPROXIMATELY four.
0548.
Due to the fact WE Have only One particular Vital Issue We are able to Possibly Suppose THIS Place Will Lessen The associated fee Perform BUT TO Confirm THIS We will Go on and USE THE Important Position AND The next Get OF PARTIAL DERIVATIVES JUST To be certain.
MEANING We will USE THIS System HERE FOR D As well as the VALUES OF THE SECOND ORDER OF PARTIAL DERIVATIVES To find out Regardless of whether Now we have A RELATIVE MAX OR MIN AT THIS Important Issue WHEN Y IS Roughly fourteen.
19 AND Z IS APPROXIMATELY 6.
08.
Listed below are The next ORDER OF PARTIALS THAT WE Observed Before.
SO We are going to BE SUBSTITUTING THIS Price FOR Y AND THIS Worth FOR Z INTO The 2nd ORDER OF PARTIALS.
WE Need to be Somewhat Very careful Even though Due to the fact Try to remember Now we have A Functionality 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 Could well be THE Z'S.
SO The 2nd Purchase OF PARTIALS WITH RESPECT TO Y IS HERE.
The 2nd Buy OF PARTIAL WITH Regard TO Z IS Listed here.
HERE'S THE Combined PARTIAL SQUARED.
Detect The way it COMES OUT To some Optimistic VALUE.
Therefore if D IS Optimistic AND SO IS The 2nd PARTIAL WITH Regard TO Y Checking out OUR NOTES Listed here Meaning We've A RELATIVE Minimum amount AT OUR Vital Position AND THEREFORE These are definitely The scale That may MINIMIZE The price of OUR BOX.
THIS WAS THE X COORDINATE From your Earlier SLIDE.
This is THE Y COORDINATE AND Here is THE Z COORDINATE WHICH Yet again ARE The size OF OUR BOX.
SO THE FRONT WIDTH WOULD BE X And that is Somewhere around 4.
05 INCHES.
THE DEPTH Could well be Y, That's Roughly 14.
19 INCHES, AND The peak Could well be Z, That's APPROXIMATELY six.
08 INCHES.
LET'S End BY Considering OUR COST Perform The place WE HAVE THE Charge Operate With regard to Y AND Z.
IN 3 Proportions THIS WOULD BE THE Area Wherever THESE Decrease AXES Might be THE Y AND Z AXIS AND THE COST Can be Together THE VERTICAL AXIS.
We will SEE THERE'S A Very low Stage HERE Which Happened AT OUR Crucial POINT THAT WE Discovered.
I HOPE YOU Identified THIS Beneficial.