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AN OPEN TOP RECTANGULAR BOX IS Currently being Made To carry A Quantity OF 350 CUBIC INCHES.
THE BASE Of your BOX IS MADE FROM Content COSTING 6 CENTS For every SQUARE INCH.
THE Entrance Of your BOX Have to be DECORATED AND WILL Charge 12 CENTS For every Sq. INCH.
The rest OF THE SIDES WILL COST two CENTS PER SQUARE INCH.
Locate The scale That should Reduce THE COST OF Developing THIS BOX.
Let us Initially DIAGRAM THE BOX AS WE SEE Below Wherever The size ARE X BY Y BY Z AND BECAUSE The amount MUST BE 350 CUBIC INCHES We've A CONSTRAINT THAT X x Y x Z Have to EQUAL 350.
BUT Prior to WE Look at OUR Price FUNCTION Allows TALK ABOUT THE SURFACE Location In the BOX.
BECAUSE THE Best IS OPEN, WE ONLY HAVE five FACES.
Let us Locate the Space With the five FACES THAT WOULD MAKE UP THE Surface area Space.
See The region OF THE FRONT Facial area Could be X x Z Which might Even be Similar to The realm Within the Again And so the Area Place HAS TWO XZ Conditions.
Detect The proper Facet OR The appropriate Deal with Might have Region Y x Z WHICH WILL BE THE Similar AS THE Remaining.
Therefore the Surface area Location Has TWO YZ Phrases AND THEN Lastly The underside HAS A place OF X x Y AND BECAUSE The highest IS OPEN WE Have only A person XY Time period Inside the SURFACE AREA AND NOW WE'LL Transform THE Area AREA TO THE COST EQUATION.
BECAUSE THE BOTTOM Price 6 CENTS PER SQUARE INCH Wherever The world OF THE BOTTOM IS X x Y Detect HOW FOR The price FUNCTION WE MULTIPLY THE XY TERM BY 6 CENTS And since THE Entrance Expenditures twelve CENTS PER SQUARE INCH Wherever THE AREA OF THE FRONT Could well be X x Z We are going to MULTIPLY THIS XZ TERM BY 12 CENTS IN THE COST Operate.
THE REMAINING SIDES Expense two CENTS For each SQUARE INCH SO THESE THREE Regions ARE ALL MULTIPLIED BY 0.
02 OR 2 CENTS.
COMBINING LIKE Phrases WE HAVE THIS Expense Purpose Listed here.
BUT NOTICE HOW We've got 3 UNKNOWNS During this EQUATION SO NOW We are going to Utilize a CONSTRAINT TO Kind A COST EQUATION WITH TWO VARIABLES.
IF WE Fix OUR CONSTRAINT FOR X BY DIVIDING BOTH SIDES BY YZ WE Will make A SUBSTITUTION FOR X INTO OUR Charge Perform Where by WE CAN SUBSTITUTE THIS FRACTION Listed here FOR X HERE AND Right here.
IF WE Do that, WE GET THIS EQUATION Listed here And when WE SIMPLIFY Discover HOW THE Element OF Z SIMPLIFIES OUT AND Listed here Issue OF Y SIMPLIFIES OUT.
SO FOR THIS FIRST TERM IF We discover THIS PRODUCT After which Go THE Y UP WE Might have 49Y For the -1 And afterwards FOR The final TERM IF WE FOUND THIS PRODUCT AND MOVED THE Z UP We might HAVE + 21Z Into the -one.
SO NOW OUR Objective IS TO MINIMIZE THIS Charge FUNCTION.
SO FOR The following Phase We are going to Locate the CRITICAL POINTS.
Crucial Details ARE Exactly where THE FUNCTION Will almost certainly HAVE MAX OR MIN Perform VALUES AND THEY Come about In which The initial Purchase OF PARTIAL DERIVATIVES ARE Both equally EQUAL TO ZERO OR WHERE Possibly Would not EXIST.
THEN When WE Discover the Essential Details, WE'LL Ascertain No matter whether Now we have A MAX OR A MIN Worth Working with OUR 2nd Get OF PARTIAL DERIVATIVES.
SO ON THIS SLIDE We are FINDING Each The initial Purchase AND SECOND Purchase OF PARTIAL DERIVATIVES.
WE Should be Somewhat Thorough Below Nevertheless Since OUR Operate Can be a Operate OF Y AND Z NOT X AND Y LIKE WE'RE Utilized to.
SO FOR The primary PARTIAL WITH RESPECT TO Y We'd DIFFERENTIATE WITH RESPECT TO Y Dealing with Z AS A continuing WHICH WOULD GIVE US THIS PARTIAL DERIVATIVE Listed here.
FOR The very first PARTIAL WITH RESPECT TO Z WE WOULD DIFFERENTIATE WITH Regard TO Z AND Handle Y AS A continuing Which might GIVE US This primary Purchase OF PARTIAL DERIVATIVE.
NOW Making use of THESE To start with Get OF PARTIAL DERIVATIVES WE CAN FIND THESE SECOND Get OF PARTIAL DERIVATIVES The place To search out THE SECOND PARTIALS WITH Regard TO Y WE WOULD DIFFERENTIATE THIS PARTIAL DERIVATIVE WITH RESPECT TO Y Once again Offering US THIS.
THE SECOND PARTIAL WITH RESPECT TO Z WE WOULD DIFFERENTIATE THIS PARTIAL By-product WITH Regard TO Z Yet again Providing US THIS.
Discover The way it'S Offered Utilizing a Damaging EXPONENT AND IN FRACTION Type After which Eventually FOR THE MIXED PARTIAL OR THE SECOND Get OF PARTIAL WITH Regard TO Y Then Z We'd DIFFERENTIATE THIS PARTIAL WITH RESPECT TO Z WHICH Detect HOW It will JUST GIVE US 0.
04.
SO NOW We'll SET The initial ORDER OF PARTIAL DERIVATIVES Equivalent TO ZERO AND Fix AS A Method OF EQUATIONS.
SO Listed here are The primary Get OF PARTIALS Established EQUAL TO ZERO.
THIS IS A FAIRLY Included Method OF EQUATIONS WHICH WE'LL Fix Making use of SUBSTITUTION.
SO I Chose to Remedy The 1st EQUATION Right here FOR Z.
SO I Additional THIS Time period TO BOTH SIDES In the EQUATION And afterwards DIVIDED BY 0.
04 Providing US THIS Benefit In this article FOR Z But when We discover THIS QUOTIENT AND MOVE Y For the -two Towards the DENOMINATOR WE CAN ALSO Produce Z AS THIS Portion HERE.
NOW THAT WE KNOW Z IS Equivalent TO THIS Portion, We will SUBSTITUTE THIS FOR Z INTO The 2nd EQUATION HERE.
WHICH IS WHAT WE SEE Listed here BUT Recognize HOW This is certainly Lifted Towards the EXPONENT OF -2 SO This could BE 1, 225 Into the -2 DIVIDED BY Y On the -4.
SO WE May take THE RECIPROCAL Which might GIVE US Y For the 4th DIVIDED BY one, five hundred, 625 AND HERE'S THE 21.
Since We now have AN EQUATION WITH Only one VARIABLE Y We wish to SOLVE THIS FOR Y.
SO FOR The initial step, There exists a Frequent Element OF Y.
SO Y = 0 WOULD Fulfill THIS EQUATION AND Will be A Essential Level BUT WE KNOW WE'RE NOT Likely To possess a DIMENSION OF ZERO SO We will JUST IGNORE THAT Price AND SET THIS EXPRESSION Below EQUAL TO ZERO AND Clear up That is WHAT WE SEE In this article.
SO WE'RE GOING TO ISOLATE THE Y CUBED Expression And afterwards CUBE ROOT Each side OF THE EQUATION.
SO IF WE Include THIS FRACTION TO Each side From the EQUATION And afterwards CHANGE THE Get In the EQUATION This is certainly WHAT WE Would've AND NOW FROM Below TO ISOLATE Y CUBED WE Really need to MULTIPLY Through the RECIPROCAL Of the FRACTION In this article.
SO Observe HOW THE Remaining SIDE SIMPLIFIES JUST Y CUBED Which Products Here's About THIS Worth HERE.
SO NOW To resolve FOR Y WE WOULD Dice ROOT Either side With the EQUATION OR RAISE BOTH SIDES From the EQUATION Towards the 1/three Electric power AND This provides Y IS Roughly 14.
1918, AND NOW TO FIND THE Z COORDINATE From the Essential Issue We will USE THIS EQUATION Right here Where by Z = 1, 225 DIVIDED BY Y SQUARED Which supplies Z IS Somewhere around 6.
0822.
WE DON'T Want IT Right this moment BUT I WENT Forward And located THE CORRESPONDING X Benefit AS WELL Utilizing OUR VOLUME FORMULA SOLVE FOR X.
SO X Can be APPROXIMATELY four.
0548.
Due to the fact WE ONLY HAVE ONE Essential Level WE CAN PROBABLY Believe THIS POINT Will Lessen The expense Purpose BUT TO VERIFY THIS We are going to Go on and USE THE Significant Issue AND The next Get OF PARTIAL DERIVATIVES JUST To be sure.
Indicating We are going to USE THIS Formulation In this article FOR D Plus the VALUES OF THE SECOND ORDER OF PARTIAL DERIVATIVES TO DETERMINE WHETHER We now have A RELATIVE MAX OR MIN AT THIS Essential POINT WHEN Y IS Close to fourteen.
19 AND Z IS Close to six.
08.
Allow me to share The 2nd ORDER OF PARTIALS THAT WE Located Previously.
SO WE'LL BE SUBSTITUTING THIS Benefit FOR Y Which VALUE FOR Z INTO THE SECOND Purchase OF PARTIALS.
WE Must be A little bit Cautious However Since REMEMBER Now we have A Operate OF Y AND Z NOT X AND Y LIKE WE NORMALLY WOULD SO THESE X'S Could be THESE Y'S AND THESE Y'S Will be THE Z'S.
SO THE SECOND Get OF PARTIALS WITH Regard TO Y IS HERE.
THE SECOND Get OF PARTIAL WITH RESPECT TO Z IS Below.
HERE'S THE Blended PARTIAL SQUARED.
Detect The way it COMES OUT To some Constructive VALUE.
SO IF D IS Optimistic AND SO IS THE SECOND PARTIAL WITH Regard TO Y Thinking about OUR NOTES Below Meaning We've got A RELATIVE MINIMUM AT OUR Significant POINT AND THEREFORE THESE ARE The size That will Limit THE COST OF OUR BOX.
THIS WAS THE X COORDINATE In the Earlier SLIDE.
HERE'S THE Y COORDINATE AND This is THE Z COORDINATE WHICH Once more ARE THE DIMENSIONS OF OUR BOX.
And so the FRONT WIDTH Can be X Which happens to be About four.
05 INCHES.
THE DEPTH Could be Y, Which happens to be Around fourteen.
19 INCHES, AND The peak Might be Z, WHICH IS Close to six.
08 INCHES.
Let us End BY LOOKING AT OUR Price Functionality In which WE Contain the Price Operate When it comes to Y AND Z.
IN 3 Proportions This could BE THE SURFACE Wherever THESE Reduce AXES WOULD BE THE Y AND Z AXIS AND The expense Will be Alongside THE VERTICAL AXIS.
WE CAN SEE THERE'S A Small Stage Listed here Which Happened AT OUR Essential Stage THAT WE Discovered.
I HOPE YOU FOUND THIS Useful.