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AN OPEN TOP RECTANGULAR BOX IS Getting CONSTRUCTED TO HOLD A Quantity OF 350 CUBIC INCHES.
THE BASE With the BOX IS Made out of Product COSTING 6 CENTS For each Sq. INCH.
THE Entrance In the BOX Needs to be DECORATED And may Expense 12 CENTS For each SQUARE INCH.
The rest OF THE SIDES WILL Expense two CENTS For each SQUARE INCH.
Come across The size Which will Lessen THE COST OF CONSTRUCTING THIS BOX.
LET'S Initially DIAGRAM THE BOX AS WE SEE Below The place THE DIMENSIONS ARE X BY Y BY Z AND BECAUSE The quantity MUST BE 350 CUBIC INCHES We have now A CONSTRAINT THAT X x Y x Z MUST EQUAL 350.
BUT Ahead of WE Discuss OUR Value Functionality LETS Look at THE Surface area Spot On the BOX.
As the TOP IS OPEN, WE ONLY HAVE five FACES.
LET'S Locate the Spot In the five FACES That will MAKE UP THE Surface area AREA.
Recognize THE AREA Of your Entrance Confront Might be X x Z WHICH WOULD Even be THE SAME AS The realm While in the Again And so the Floor Space HAS TWO XZ Phrases.
Recognize THE RIGHT Facet OR THE RIGHT Experience WOULD HAVE Location Y x Z WHICH WILL BE THE Similar As being the Remaining.
Hence the SURFACE Spot CONTAINS TWO YZ Conditions AND THEN Ultimately The underside HAS A place OF X x Y And since The highest IS OPEN WE Have only One particular XY TERM Inside the Area Space AND NOW WE'LL CONVERT THE Floor Space TO THE COST EQUATION.
BECAUSE THE Base Value 6 CENTS For each SQUARE INCH Wherever The world OF THE BOTTOM IS X x Y Detect HOW FOR The expense Operate WE MULTIPLY THE XY Phrase BY six CENTS AND BECAUSE THE Entrance Charges 12 CENTS For each Sq. INCH The place The region With the FRONT Will be X x Z We will MULTIPLY THIS XZ Expression BY twelve CENTS IN The fee Operate.
THE REMAINING SIDES Value 2 CENTS For every SQUARE INCH SO THESE 3 Places ARE ALL MULTIPLIED BY 0.
02 OR 2 CENTS.
COMBINING LIKE Conditions Now we have THIS Value Purpose Right here.
BUT Observe HOW We have now A few UNKNOWNS IN THIS EQUATION SO NOW We are going to Make use of a CONSTRAINT TO Kind A COST EQUATION WITH TWO VARIABLES.
IF WE Clear up OUR CONSTRAINT FOR X BY DIVIDING Either side BY YZ WE CAN MAKE A SUBSTITUTION FOR X INTO OUR Price Operate WHERE We can easily SUBSTITUTE THIS FRACTION In this article FOR X HERE AND Below.
IF WE DO THIS, WE GET THIS EQUATION HERE And when WE SIMPLIFY NOTICE HOW THE Component OF Z SIMPLIFIES OUT AND Below FACTOR OF Y SIMPLIFIES OUT.
SO FOR THIS FIRST Time period IF We discover THIS Products After which MOVE THE Y UP WE Would've 49Y For the -1 After which you can FOR THE LAST Phrase IF WE Located THIS Products AND MOVED THE Z UP WE'D HAVE + 21Z For the -one.
SO NOW OUR Target IS TO MINIMIZE THIS Expense Purpose.
SO FOR The following Phase We are going to Locate the Significant Details.
Essential POINTS ARE Where by THE Purpose IS GOING TO HAVE MAX OR MIN Perform VALUES They usually OCCUR Wherever The primary Get OF PARTIAL DERIVATIVES ARE The two Equivalent TO ZERO OR In which EITHER Won't EXIST.
THEN ONCE WE FIND THE Essential Factors, We are going to Identify No matter if Now we have A MAX Or even a MIN Price Utilizing OUR 2nd ORDER OF PARTIAL DERIVATIVES.
SO ON THIS SLIDE WE'RE Getting Both of those The initial ORDER AND 2nd ORDER OF PARTIAL DERIVATIVES.
WE Need to be A bit Very careful In this article While For the reason that OUR Perform Is usually a Functionality OF Y AND Z NOT X AND Y LIKE WE'RE USED TO.
SO FOR THE FIRST PARTIAL WITH Regard TO Y We'd DIFFERENTIATE WITH Regard TO Y Dealing with Z AS A CONSTANT WHICH WOULD GIVE US THIS PARTIAL DERIVATIVE Below.
FOR The initial PARTIAL WITH RESPECT TO Z WE WOULD DIFFERENTIATE WITH Regard TO Z AND TREAT Y AS A continuing WHICH WOULD GIVE US This primary Buy OF PARTIAL Spinoff.
NOW USING THESE Initially Get OF PARTIAL DERIVATIVES WE Can discover THESE 2nd Get OF PARTIAL DERIVATIVES In which To seek out The 2nd PARTIALS WITH Regard TO Y We'd DIFFERENTIATE THIS PARTIAL Spinoff WITH Regard TO Y AGAIN Providing US THIS.
THE SECOND PARTIAL WITH Regard TO Z We'd DIFFERENTIATE THIS PARTIAL Spinoff WITH RESPECT TO Z Yet again GIVING US THIS.
Discover The way it'S Supplied USING A NEGATIVE EXPONENT AND IN Portion Type After which you can At last For your MIXED PARTIAL OR THE SECOND Purchase OF PARTIAL WITH RESPECT TO Y And afterwards Z WE WOULD DIFFERENTIATE THIS PARTIAL WITH Regard TO Z WHICH NOTICE HOW It will JUST GIVE US 0.
04.
SO NOW We will SET THE FIRST Buy OF PARTIAL DERIVATIVES EQUAL TO ZERO AND SOLVE As being a Procedure OF EQUATIONS.
SO Here's The primary Get OF PARTIALS SET EQUAL TO ZERO.
THIS Is a reasonably Included SYSTEM OF EQUATIONS WHICH WE'LL Resolve Working with SUBSTITUTION.
SO I DECIDED TO Address The primary EQUATION In this article FOR Z.
SO I Extra THIS Expression TO Either side Of your EQUATION After which you can DIVIDED BY 0.
04 Supplying US THIS VALUE Listed here FOR Z However, if WE FIND THIS QUOTIENT AND Shift Y To your -2 To your DENOMINATOR WE CAN ALSO WRITE Z AS THIS Portion HERE.
Given that We all know Z IS EQUAL TO THIS Portion, We can easily SUBSTITUTE THIS FOR Z INTO The 2nd EQUATION Right here.
Which happens to be WHAT WE SEE HERE BUT Recognize HOW This is certainly RAISED Into the EXPONENT OF -two SO This may BE 1, 225 For the -2 DIVIDED BY Y Towards the -4.
SO WE Might take THE RECIPROCAL WHICH WOULD GIVE US Y On the 4th DIVIDED BY 1, five hundred, 625 AND This is THE 21.
NOW THAT We've AN EQUATION WITH Only one VARIABLE Y WE WANT TO Resolve THIS FOR Y.
SO FOR THE FIRST STEP, You will find there's COMMON FACTOR OF Y.
SO Y = 0 WOULD SATISFY THIS EQUATION AND Can be A Important Position BUT WE KNOW WE'RE NOT GOING To possess a DIMENSION OF ZERO SO We will JUST Dismiss THAT Benefit AND SET THIS EXPRESSION HERE Equivalent TO ZERO AND Clear up Which can be WHAT WE SEE HERE.
SO We'll ISOLATE THE Y CUBED Expression Then Dice ROOT Either side From the EQUATION.
SO IF WE ADD THIS Portion TO BOTH SIDES From the EQUATION After which you can CHANGE THE Buy From the EQUATION This is often WHAT WE Would've AND NOW FROM Below TO ISOLATE Y CUBED WE Really need to MULTIPLY Because of the RECIPROCAL OF THIS FRACTION In this article.
SO Discover HOW THE LEFT SIDE SIMPLIFIES JUST Y CUBED AND THIS PRODUCT HERE IS APPROXIMATELY THIS Worth Below.
SO NOW To unravel FOR Y WE WOULD CUBE ROOT BOTH SIDES In the EQUATION OR RAISE BOTH SIDES From the EQUATION On the 1/3 Energy AND This offers Y IS Close to 14.
1918, AND NOW TO Locate the Z COORDINATE With the Essential Stage We are able to USE THIS EQUATION HERE The place Z = one, 225 DIVIDED BY Y SQUARED Which supplies Z IS Somewhere around 6.
0822.
WE DON'T Have to have IT Today BUT I WENT In advance And located THE CORRESPONDING X Worth In addition Employing OUR Quantity FORMULA Remedy FOR X.
SO X Could well be Around four.
0548.
Since WE ONLY HAVE ONE Significant Place We will Almost certainly Presume THIS Position Will probably Decrease The expense Functionality BUT TO VERIFY THIS We will GO AHEAD AND Make use of the Vital Level AND THE SECOND Get OF PARTIAL DERIVATIVES JUST To make certain.
That means We will USE THIS System In this article FOR D And also the VALUES OF THE SECOND ORDER OF PARTIAL DERIVATIVES To find out Regardless of whether We have now A RELATIVE MAX OR MIN AT THIS CRITICAL Place WHEN Y IS Somewhere around fourteen.
19 AND Z IS About six.
08.
HERE ARE THE SECOND Get OF PARTIALS THAT WE FOUND EARLIER.
SO WE'LL BE SUBSTITUTING THIS Price FOR Y Which Price FOR Z INTO The 2nd Get OF PARTIALS.
WE Must be Just a little Very careful However Since Recall WE HAVE A Functionality OF Y AND Z NOT X AND Y LIKE WE Typically WOULD SO THESE X'S Will be THESE Y'S AND THESE Y'S Might be THE Z'S.
SO THE SECOND ORDER OF PARTIALS WITH RESPECT TO Y IS Right here.
THE SECOND Buy OF PARTIAL WITH Regard TO Z IS HERE.
Here is THE Blended PARTIAL SQUARED.
Detect HOW IT Arrives OUT TO A POSITIVE Price.
SO IF D IS POSITIVE AND SO IS The 2nd PARTIAL WITH Regard TO Y Taking a look at OUR NOTES Listed here Which means We now have A RELATIVE Minimum amount AT OUR Essential Stage And for that reason These are definitely The scale That could MINIMIZE THE COST OF OUR BOX.
THIS WAS THE X COORDINATE From your Former SLIDE.
This is THE Y COORDINATE AND This is THE Z COORDINATE WHICH All over again ARE THE DIMENSIONS OF OUR BOX.
SO THE FRONT WIDTH Can be X That's Roughly four.
05 INCHES.
THE DEPTH Will be Y, That's Around 14.
19 INCHES, AND THE HEIGHT Could well be Z, Which happens to be Around six.
08 INCHES.
LET'S Complete BY Thinking about OUR Charge Perform The place WE Hold the COST Perform With regard to Y AND Z.
IN A few Proportions This is able to BE THE SURFACE Exactly where THESE Reduce AXES Might be THE Y AND Z AXIS AND The price Could well be Together THE VERTICAL AXIS.
We will SEE There is a Small Stage Below AND THAT Happened AT OUR Important Stage THAT WE Observed.
I HOPE YOU Observed THIS Practical.