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AN OPEN Major RECTANGULAR BOX IS Staying Built TO HOLD A Quantity OF 350 CUBIC INCHES.

THE BASE On the BOX IS Made out of Substance COSTING six CENTS PER Sq. INCH.

THE Entrance In the BOX Have to be DECORATED AND WILL Price tag 12 CENTS PER Sq. INCH.

The rest OF THE SIDES WILL Expense two CENTS For every Sq. INCH.

Obtain THE DIMENSIONS That may Lessen THE COST OF Setting up THIS BOX.

Let us Initial DIAGRAM THE BOX AS WE SEE In this article WHERE The scale ARE X BY Y BY Z And since The amount MUST BE 350 CUBIC INCHES We've got A CONSTRAINT THAT X x Y x Z Have to EQUAL 350.

BUT Prior to WE TALK ABOUT OUR Value Operate LETS Take a look at THE Area Spot On the BOX.

As the Best IS OPEN, WE Have only 5 FACES.

LET'S FIND THE Location OF THE five FACES THAT WOULD MAKE UP THE SURFACE Spot.

Observe The realm In the Entrance Experience Will be X x Z WHICH WOULD Even be THE SAME AS The world IN THE Again SO THE Floor Region HAS TWO XZ Conditions.

Recognize The proper Facet OR The ideal Deal with Might have AREA Y x Z WHICH Would be the Exact As being the LEFT.

Therefore the SURFACE Space Incorporates TWO YZ Phrases After which Last but not least The underside HAS AN AREA OF X x Y AND BECAUSE The best IS OPEN WE ONLY HAVE A single XY Time period While in the Area Region AND NOW WE'LL CONVERT THE Area AREA TO The fee EQUATION.

As the BOTTOM Price 6 CENTS For every Sq. INCH In which The world OF The underside IS X x Y See HOW FOR The price Purpose WE MULTIPLY THE XY Phrase BY 6 CENTS And since THE FRONT Expenses 12 CENTS PER SQUARE INCH Wherever The world Of your FRONT Will be X x Z We are going to MULTIPLY THIS XZ TERM BY 12 CENTS IN The associated fee Purpose.

THE REMAINING SIDES Expense 2 CENTS For each Sq. INCH SO THESE THREE Regions ARE ALL MULTIPLIED BY 0.

02 OR 2 CENTS.

COMBINING LIKE Phrases Now we have THIS Value Purpose Listed here.

BUT Discover HOW We've THREE UNKNOWNS With this EQUATION SO NOW We will USE A CONSTRAINT TO Type A price EQUATION WITH TWO VARIABLES.

IF WE Address OUR CONSTRAINT FOR X BY DIVIDING Either side BY YZ WE Will make A SUBSTITUTION FOR X INTO OUR Price tag Operate Exactly where We are able to SUBSTITUTE THIS Portion Listed here FOR X In this article AND In this article.

IF WE DO THIS, WE GET THIS EQUATION In this article And when WE SIMPLIFY Observe HOW THE FACTOR OF Z SIMPLIFIES OUT AND Right here Component OF Y SIMPLIFIES OUT.

SO FOR THIS FIRST TERM IF WE FIND THIS Products Then Go THE Y UP WE Might have 49Y Towards the -1 And afterwards FOR THE LAST Expression IF WE Identified THIS PRODUCT AND MOVED THE Z UP We might HAVE + 21Z Into the -one.

SO NOW OUR Target IS To reduce THIS Value FUNCTION.

SO FOR Another Stage We will Locate the Significant Factors.

Essential Details ARE In which THE Purpose Will almost certainly HAVE MAX OR MIN Operate VALUES Plus they Manifest WHERE The primary Get OF PARTIAL DERIVATIVES ARE BOTH EQUAL TO ZERO OR The place Both Would not EXIST.

THEN After WE Discover the Significant Factors, We are going to Figure out Irrespective of whether WE HAVE A MAX Or maybe a MIN Price Working with OUR 2nd Purchase OF PARTIAL DERIVATIVES.

SO ON THIS SLIDE We are Obtaining Each The 1st Get AND Next Buy OF PARTIAL DERIVATIVES.

WE Ought to be A LITTLE Mindful Right here While BECAUSE OUR Functionality Is really a Functionality OF Y AND Z NOT X AND Y LIKE We are USED TO.

SO FOR THE FIRST PARTIAL WITH Regard TO Y We might DIFFERENTIATE WITH Regard TO Y Managing Z AS A relentless WHICH WOULD GIVE US THIS PARTIAL Spinoff Listed here.

FOR The 1st PARTIAL WITH RESPECT TO Z We might DIFFERENTIATE WITH RESPECT TO Z AND Take care of Y AS A CONSTANT Which might GIVE US THIS FIRST Get OF PARTIAL Spinoff.

NOW Utilizing THESE Very first Get OF PARTIAL DERIVATIVES WE Can discover THESE SECOND Buy OF PARTIAL DERIVATIVES Exactly where To seek out The 2nd PARTIALS WITH Regard TO Y We'd DIFFERENTIATE THIS PARTIAL By-product WITH RESPECT TO Y AGAIN Offering US THIS.

The 2nd PARTIAL WITH Regard TO Z WE WOULD DIFFERENTIATE THIS PARTIAL DERIVATIVE WITH Regard TO Z Once more Providing US THIS.

NOTICE HOW IT'S Supplied Employing a NEGATIVE EXPONENT AND IN Portion Kind Then Ultimately For your MIXED PARTIAL OR The next Get OF PARTIAL WITH RESPECT TO Y After which you can Z We might DIFFERENTIATE THIS PARTIAL WITH RESPECT TO Z WHICH Observe HOW It might JUST GIVE US 0.

04.

SO NOW We will SET THE FIRST ORDER OF PARTIAL DERIVATIVES EQUAL TO ZERO AND Clear up Being a SYSTEM OF EQUATIONS.

SO Listed below are The very first ORDER OF PARTIALS SET EQUAL TO ZERO.

THIS Is a reasonably Concerned Technique OF EQUATIONS WHICH WE'LL Remedy Making use of SUBSTITUTION.

SO I DECIDED TO SOLVE The very first EQUATION HERE FOR Z.

SO I Included THIS Phrase TO BOTH SIDES On the EQUATION And afterwards DIVIDED BY 0.

04 Providing US THIS Benefit Right here FOR Z BUT IF WE FIND THIS QUOTIENT AND Shift Y Into the -two Towards the DENOMINATOR WE Could also Publish Z AS THIS FRACTION Listed here.

Since WE KNOW Z IS EQUAL TO THIS FRACTION, WE CAN SUBSTITUTE THIS FOR Z INTO The 2nd EQUATION Below.

That's WHAT WE SEE Below BUT Detect HOW That is Lifted On the EXPONENT OF -2 SO This could BE 1, 225 Towards the -two DIVIDED BY Y Into the -4.

SO WE Usually takes THE RECIPROCAL Which might GIVE US Y On the 4th DIVIDED BY one, 500, 625 AND This is THE 21.

Since Now we have AN EQUATION WITH JUST ONE VARIABLE Y WE WANT TO Fix THIS FOR Y.

SO FOR The initial step, THERE IS A Widespread Element OF Y.

SO Y = 0 WOULD SATISFY THIS EQUATION AND Might be A Essential POINT BUT WE KNOW We are NOT GOING To possess a DIMENSION OF ZERO SO WE'LL JUST Overlook THAT Worth AND Established THIS EXPRESSION Below Equivalent TO ZERO AND Clear up WHICH IS WHAT WE SEE Below.

SO We will ISOLATE THE Y CUBED TERM And afterwards CUBE ROOT Each side In the EQUATION.

SO IF WE Include THIS FRACTION TO Either side OF THE EQUATION And after that CHANGE THE Get On the EQUATION This is often WHAT WE Would've AND NOW FROM Right here TO ISOLATE Y CUBED WE Must MULTIPLY With the RECIPROCAL Of the Portion Right here.

SO Observe HOW THE Remaining Aspect SIMPLIFIES JUST Y CUBED AND THIS Products HERE IS Close to THIS VALUE Below.

SO NOW To resolve FOR Y We might CUBE ROOT Either side Of your EQUATION OR Increase Each side From the EQUATION For the one/three Energy AND THIS GIVES Y IS About fourteen.

1918, AND NOW TO FIND THE Z COORDINATE On the Important Place We will USE THIS EQUATION Right here In which Z = one, 225 DIVIDED BY Y SQUARED Which supplies Z IS Close to 6.

0822.

WE DON'T NEED IT At the moment BUT I WENT AHEAD And located THE CORRESPONDING X Worth At the same time Making use of OUR Quantity Formulation Clear up FOR X.

SO X Can be APPROXIMATELY 4.

0548.

Mainly because WE ONLY HAVE 1 Significant Place WE CAN Almost certainly ASSUME THIS Place Will almost certainly Reduce The fee FUNCTION BUT TO Validate THIS We will Go on and USE THE CRITICAL POINT AND The 2nd ORDER OF PARTIAL DERIVATIVES JUST To ensure.

This means WE'LL USE THIS Components Listed here FOR D Along with the VALUES OF The 2nd Buy OF PARTIAL DERIVATIVES To find out Whether or not Now we have A RELATIVE MAX OR MIN AT THIS Essential Level WHEN Y IS About 14.

19 AND Z IS Roughly six.

08.

HERE ARE The 2nd Buy OF PARTIALS THAT WE Discovered Previously.

SO We will BE SUBSTITUTING THIS VALUE FOR Y AND THIS Worth FOR Z INTO The 2nd Get OF PARTIALS.

WE Need to be A LITTLE Watchful However Since Keep in mind We've got A Perform 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 Could be THE Z'S.

SO The next Buy OF PARTIALS WITH Regard TO Y IS HERE.

The 2nd Purchase OF PARTIAL WITH Regard TO Z IS Listed here.

HERE'S THE MIXED PARTIAL SQUARED.

Detect HOW IT COMES OUT To some Favourable Benefit.

Therefore if D IS Optimistic AND SO IS THE SECOND PARTIAL WITH Regard TO Y LOOKING AT OUR NOTES In this article Meaning We have now A RELATIVE Bare minimum AT OUR Important Level And thus They are THE DIMENSIONS That may MINIMIZE The expense of OUR BOX.

THIS WAS THE X COORDINATE Within the Earlier SLIDE.

HERE'S THE Y COORDINATE AND This is THE Z COORDINATE WHICH Once more ARE The size OF OUR BOX.

Hence the Entrance WIDTH Could well be X That's About 4.

05 INCHES.

THE DEPTH Could well be Y, That's Roughly 14.

19 INCHES, AND THE HEIGHT Can be Z, Which happens to be Close to 6.

08 INCHES.

LET'S End BY Thinking about OUR Charge Operate WHERE WE Contain the Value Perform IN TERMS OF Y AND Z.

IN 3 Proportions This could BE THE SURFACE Where by THESE Lessen AXES WOULD BE THE Y AND Z AXIS AND The fee Can be Alongside THE VERTICAL AXIS.

WE CAN SEE THERE'S A Reduced Stage Listed here Which OCCURRED AT OUR Crucial Place THAT WE Uncovered.

I HOPE YOU Located THIS Practical.