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

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AN Open up Leading RECTANGULAR BOX IS Getting CONSTRUCTED To carry A Quantity OF 350 CUBIC INCHES.

THE BASE Of your BOX IS Made out of Content COSTING 6 CENTS For each SQUARE INCH.

THE Entrance From the BOX Should be DECORATED And can COST 12 CENTS PER SQUARE INCH.

THE REMAINDER OF The perimeters WILL Expense two CENTS For every SQUARE INCH.

Locate The size That could Decrease THE COST OF CONSTRUCTING THIS BOX.

LET'S Initially DIAGRAM THE BOX AS WE SEE HERE Exactly where The size ARE X BY Y BY Z AND BECAUSE The amount Has to be 350 CUBIC INCHES We've got A CONSTRAINT THAT X x Y x Z Need to Equivalent 350.

BUT Right before WE Take a look at OUR Price tag Functionality LETS Mention THE Floor Region In the BOX.

Since the TOP IS Open up, WE Have only five FACES.

LET'S Discover the Place On the 5 FACES That may MAKE UP THE SURFACE AREA.

Discover THE AREA From the FRONT FACE Might be X x Z Which might Even be Similar to THE AREA During the Back again Hence the Area Space HAS TWO XZ TERMS.

NOTICE The proper SIDE OR THE RIGHT Deal with Might have AREA Y x Z WHICH Would be the Very same Because the LEFT.

Hence the Floor Place Includes TWO YZ Phrases After which you can Eventually The underside HAS A region OF X x Y And since THE TOP IS OPEN WE Have only A single XY Time period From the Area Space AND NOW We are going to Change THE Floor Place TO The associated fee EQUATION.

Since the Base Charge six CENTS For each Sq. INCH Where by THE AREA OF THE BOTTOM IS X x Y Detect HOW FOR The fee Perform WE MULTIPLY THE XY TERM BY six CENTS And since THE FRONT Charges twelve CENTS PER SQUARE INCH In which The region Of your FRONT Can be X x Z We are going to MULTIPLY THIS XZ Time period BY 12 CENTS IN THE COST Operate.

THE REMAINING SIDES Charge two CENTS For every Sq. INCH SO THESE THREE Regions ARE ALL MULTIPLIED BY 0.

02 OR two CENTS.

COMBINING LIKE Conditions We've THIS COST FUNCTION Listed here.

BUT Detect HOW We now have THREE UNKNOWNS Within this EQUATION SO NOW We will Utilize a CONSTRAINT TO Type A value EQUATION WITH TWO VARIABLES.

IF WE Resolve OUR CONSTRAINT FOR X BY DIVIDING Each side BY YZ WE Might make A SUBSTITUTION FOR X INTO OUR Value Perform The place We can easily SUBSTITUTE THIS FRACTION Listed here FOR X Listed here AND Below.

IF WE Make this happen, WE GET THIS EQUATION HERE And when WE SIMPLIFY See HOW THE Component OF Z SIMPLIFIES OUT AND Right here FACTOR OF Y SIMPLIFIES OUT.

SO FOR This primary Time period IF We discover THIS Product or service Then MOVE THE Y UP WE WOULD HAVE 49Y Into the -1 And after that FOR THE LAST Phrase IF WE Discovered THIS Item AND MOVED THE Z UP We might HAVE + 21Z On the -1.

SO NOW OUR Objective IS TO MINIMIZE THIS Price Purpose.

SO FOR The subsequent Action We will Discover the Essential Factors.

Crucial Factors ARE Wherever THE Purpose Will probably HAVE MAX OR MIN Functionality VALUES Plus they Manifest In which The main Buy OF PARTIAL DERIVATIVES ARE Both of those EQUAL TO ZERO OR The place Both Will not EXIST.

THEN The moment WE FIND THE CRITICAL POINTS, We will DETERMINE Regardless of whether We now have A MAX Or maybe a MIN Worth USING OUR 2nd ORDER OF PARTIAL DERIVATIVES.

SO ON THIS SLIDE WE'RE FINDING BOTH THE FIRST Get AND Next Purchase OF PARTIAL DERIVATIVES.

WE Need to be Somewhat Watchful Listed here While Since OUR Perform Is often a Purpose OF Y AND Z NOT X AND Y LIKE WE'RE USED TO.

SO FOR The primary PARTIAL WITH RESPECT TO Y WE WOULD DIFFERENTIATE WITH RESPECT TO Y TREATING Z AS A continuing WHICH WOULD GIVE US THIS PARTIAL Spinoff Right here.

FOR The main PARTIAL WITH RESPECT TO Z We might DIFFERENTIATE WITH RESPECT TO Z AND Address Y AS A CONSTANT Which might GIVE US THIS FIRST Get OF PARTIAL By-product.

NOW Applying THESE 1st Get OF PARTIAL DERIVATIVES WE Can discover THESE SECOND ORDER OF PARTIAL DERIVATIVES The place To seek out The next PARTIALS WITH RESPECT TO Y We'd DIFFERENTIATE THIS PARTIAL Spinoff WITH Regard TO Y Once again Providing US THIS.

The next PARTIAL WITH RESPECT TO Z WE WOULD DIFFERENTIATE THIS PARTIAL DERIVATIVE WITH Regard TO Z Yet again Offering US THIS.

Detect The way it'S Presented USING A Detrimental EXPONENT As well as in Portion Sort Then Lastly With the MIXED PARTIAL OR THE SECOND Buy OF PARTIAL WITH Regard TO Y And after that Z WE WOULD DIFFERENTIATE THIS PARTIAL WITH Regard TO Z WHICH Observe HOW It could JUST GIVE US 0.

04.

SO NOW We'll Established The primary Purchase OF PARTIAL DERIVATIVES EQUAL TO ZERO AND Resolve Being a Program OF EQUATIONS.

SO Here's The 1st Get OF PARTIALS Established Equivalent TO ZERO.

THIS IS A FAIRLY Included Method OF EQUATIONS WHICH We will Address Making use of SUBSTITUTION.

SO I DECIDED TO Address The 1st EQUATION Listed here FOR Z.

SO I Additional THIS Expression TO Either side From the EQUATION After which you can DIVIDED BY 0.

04 Offering US THIS Benefit In this article FOR Z But when WE FIND THIS QUOTIENT AND Shift Y Into the -two TO THE DENOMINATOR WE May also Generate Z AS THIS FRACTION In this article.

NOW THAT We all know Z IS Equivalent TO THIS FRACTION, WE CAN SUBSTITUTE THIS FOR Z INTO THE SECOND EQUATION In this article.

Which happens to be WHAT WE SEE Listed here BUT See HOW THIS IS Elevated On the EXPONENT OF -two SO THIS WOULD BE 1, 225 Towards the -two DIVIDED BY Y For the -four.

SO WE Usually takes THE RECIPROCAL Which might GIVE US Y Into the 4th DIVIDED BY one, five hundred, 625 AND Here is THE 21.

NOW THAT We've got AN EQUATION WITH Only one VARIABLE Y WE WANT TO SOLVE THIS FOR Y.

SO FOR THE FIRST STEP, THERE IS A COMMON Component OF Y.

SO Y = 0 WOULD Fulfill THIS EQUATION AND Might be A CRITICAL Position BUT WE KNOW WE'RE NOT GOING TO HAVE A DIMENSION OF ZERO SO We are going to JUST Overlook THAT Benefit AND Established THIS EXPRESSION In this article Equivalent TO ZERO AND Fix Which happens to be WHAT WE SEE Below.

SO We will ISOLATE THE Y CUBED Phrase And after that CUBE ROOT Either side Of your EQUATION.

Therefore if WE Incorporate THIS FRACTION TO Each side OF THE EQUATION After which Alter the ORDER From the EQUATION This can be WHAT WE WOULD HAVE AND NOW FROM HERE TO ISOLATE Y CUBED WE Really need to MULTIPLY BY THE RECIPROCAL Of the Portion Below.

SO Observe HOW THE Remaining Facet SIMPLIFIES JUST Y CUBED AND THIS Merchandise Here's Roughly THIS Price HERE.

SO NOW To resolve FOR Y WE WOULD Dice ROOT Each side Of your EQUATION OR RAISE Either side With the EQUATION For the 1/3 Electric power AND This offers Y IS Somewhere around 14.

1918, AND NOW TO Discover the Z COORDINATE With the Vital Position We could USE THIS EQUATION In this article Wherever Z = one, 225 DIVIDED BY Y SQUARED Which provides Z IS Somewhere around six.

0822.

We do not NEED IT At this moment BUT I WENT AHEAD And located THE CORRESPONDING X VALUE In addition Applying OUR Quantity Method Fix FOR X.

SO X Can be Around 4.

0548.

BECAUSE WE Have only One particular Significant Place We can easily Possibly ASSUME THIS Issue Will Lower The price Purpose BUT TO Validate THIS WE'LL Go on and Make use of the CRITICAL Issue AND The next Purchase OF PARTIAL DERIVATIVES JUST To verify.

This means We will USE THIS FORMULA Right here FOR D As well as VALUES OF The 2nd Buy OF PARTIAL DERIVATIVES To find out WHETHER WE HAVE A RELATIVE MAX OR MIN AT THIS CRITICAL Level WHEN Y IS About 14.

19 AND Z IS APPROXIMATELY 6.

08.

Here's The 2nd Get OF PARTIALS THAT WE Observed EARLIER.

SO We will BE SUBSTITUTING THIS VALUE FOR Y Which Worth FOR Z INTO THE SECOND Buy OF PARTIALS.

WE Needs to be A LITTLE Watchful While BECAUSE Don't forget We've A Functionality OF Y AND Z NOT X AND Y LIKE WE Usually WOULD SO THESE X'S Might be THESE Y'S AND THESE Y'S Can be THE Z'S.

SO THE SECOND Buy OF PARTIALS WITH RESPECT TO Y IS Right here.

The next Get OF PARTIAL WITH RESPECT TO Z IS HERE.

HERE'S THE MIXED PARTIAL SQUARED.

Recognize HOW IT COMES OUT To the Optimistic Price.

Therefore if D IS Constructive AND SO IS THE SECOND PARTIAL WITH Regard TO Y Investigating OUR NOTES Below THAT MEANS We've got A RELATIVE Bare minimum AT OUR Vital Stage AND THEREFORE These are typically The scale THAT WOULD Lower THE COST OF OUR BOX.

THIS WAS THE X COORDINATE In the Former SLIDE.

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

Therefore the FRONT WIDTH WOULD BE X Which happens to be About four.

05 INCHES.

THE DEPTH Might be Y, That is Close to fourteen.

19 INCHES, AND The peak Will be Z, That is APPROXIMATELY 6.

08 INCHES.

Let us FINISH BY Considering OUR Price tag Functionality WHERE WE Provide the Price Perform With regard to Y AND Z.

IN THREE Proportions THIS WOULD BE THE SURFACE Where by THESE Reduce AXES Will be THE Y AND Z AXIS AND THE COST Might be Together THE VERTICAL AXIS.

WE CAN SEE There is a Small Place Listed here Which OCCURRED AT OUR Vital Level THAT WE Uncovered.

I HOPE YOU FOUND THIS Handy.