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AN OPEN Top rated RECTANGULAR BOX IS Staying Made TO HOLD A VOLUME OF 350 CUBIC INCHES.
The bottom OF THE BOX IS MADE FROM Product COSTING six CENTS For each Sq. INCH.
THE FRONT On the BOX Needs to be DECORATED AND WILL Charge twelve CENTS PER SQUARE INCH.
The rest OF THE SIDES WILL Price tag two CENTS For every Sq. INCH.
Discover The size THAT WILL Lower The price of CONSTRUCTING THIS BOX.
Let us Initially DIAGRAM THE BOX AS WE SEE Below WHERE THE DIMENSIONS ARE X BY Y BY Z AND BECAUSE The amount Should be 350 CUBIC INCHES We've A CONSTRAINT THAT X x Y x Z Will have to Equivalent 350.
BUT Just before WE Look at OUR Price tag FUNCTION LETS Take a look at THE Floor AREA In the BOX.
Since the Top rated IS OPEN, WE ONLY HAVE five FACES.
LET'S Discover the Spot On the 5 FACES That may MAKE UP THE Area Space.
Recognize The realm From the Entrance Deal with WOULD BE X x Z Which might Even be Similar to The region During the Back again Hence the Surface area Place HAS TWO XZ Phrases.
NOTICE The ideal Aspect OR The appropriate Encounter Might have Space Y x Z WHICH WILL BE THE Exact same Because the Remaining.
And so the SURFACE Place Incorporates TWO YZ TERMS And afterwards Last but not least THE BOTTOM HAS A location OF X x Y And since The very best IS OPEN WE Have only A single XY Phrase From the Area Location AND NOW WE'LL CONVERT THE SURFACE Spot TO THE COST EQUATION.
Since the BOTTOM Value 6 CENTS For every Sq. INCH In which The region OF The underside IS X x Y Recognize HOW FOR The fee Functionality WE MULTIPLY THE XY Time period BY 6 CENTS And since THE Entrance Charges 12 CENTS For every SQUARE INCH The place The world With the Entrance Will be X x Z We are going to MULTIPLY THIS XZ Time period BY twelve CENTS IN The expense Perform.
THE REMAINING SIDES COST two CENTS PER Sq. INCH SO THESE THREE Places ARE ALL MULTIPLIED BY 0.
02 OR 2 CENTS.
COMBINING LIKE Phrases We have now THIS Expense FUNCTION HERE.
BUT Detect HOW We have now THREE UNKNOWNS In this particular EQUATION SO NOW We will Make use of a CONSTRAINT TO Variety A value EQUATION WITH TWO VARIABLES.
IF WE Remedy OUR CONSTRAINT FOR X BY DIVIDING BOTH SIDES BY YZ WE CAN MAKE A SUBSTITUTION FOR X INTO OUR Expense Operate Where by We could SUBSTITUTE THIS Portion In this article FOR X In this article AND HERE.
IF WE Try this, WE GET THIS EQUATION In this article AND IF WE SIMPLIFY Discover HOW THE Variable OF Z SIMPLIFIES OUT AND Below Variable OF Y SIMPLIFIES OUT.
SO FOR THIS FIRST Phrase IF WE FIND THIS PRODUCT And after that Transfer THE Y UP WE WOULD HAVE 49Y On the -one AND THEN FOR The final Expression IF WE FOUND THIS Solution AND MOVED THE Z UP We might HAVE + 21Z For the -one.
SO NOW OUR Intention IS To reduce THIS Charge Perform.
SO FOR THE NEXT Action We will Locate the Important Factors.
Essential POINTS ARE In which THE Operate Will HAVE MAX OR MIN Functionality VALUES Plus they Happen The place The very first Purchase OF PARTIAL DERIVATIVES ARE Both of those Equivalent TO ZERO OR In which EITHER DOES NOT EXIST.
THEN The moment WE FIND THE CRITICAL Details, We are going to Establish Regardless of whether We now have A MAX Or even a MIN Worth Applying OUR 2nd Get OF PARTIAL DERIVATIVES.
SO ON THIS SLIDE We are Acquiring Each The initial ORDER AND 2nd Purchase OF PARTIAL DERIVATIVES.
WE Must be Somewhat Cautious In this article Even though BECAUSE OUR Perform IS A Operate OF Y AND Z NOT X AND Y LIKE WE'RE Accustomed to.
SO FOR The initial PARTIAL WITH RESPECT TO Y WE WOULD DIFFERENTIATE WITH Regard TO Y Dealing with Z AS A continuing Which might GIVE US THIS PARTIAL By-product HERE.
FOR The 1st PARTIAL WITH Regard TO Z WE WOULD DIFFERENTIATE WITH Regard TO Z AND TREAT Y AS A CONSTANT WHICH WOULD GIVE US This primary Buy OF PARTIAL DERIVATIVE.
NOW Making use of THESE 1st Get OF PARTIAL DERIVATIVES WE Can discover THESE Next ORDER OF PARTIAL DERIVATIVES WHERE To locate The next PARTIALS WITH Regard TO Y WE WOULD DIFFERENTIATE THIS PARTIAL DERIVATIVE WITH RESPECT TO Y Once more Supplying US THIS.
The next PARTIAL WITH RESPECT TO Z We might DIFFERENTIATE THIS PARTIAL Spinoff WITH RESPECT TO Z Yet again Providing US THIS.
Observe HOW IT'S Presented Utilizing a Adverse EXPONENT AND IN Portion FORM And afterwards Last but not least For that Combined PARTIAL OR The next ORDER OF PARTIAL WITH RESPECT TO Y Then Z We'd DIFFERENTIATE THIS PARTIAL WITH Regard TO Z WHICH Recognize HOW It might JUST GIVE US 0.
04.
SO NOW We will Established The main Purchase OF PARTIAL DERIVATIVES Equivalent TO ZERO AND Clear up AS A Program OF EQUATIONS.
SO HERE ARE The initial Get OF PARTIALS SET EQUAL TO ZERO.
THIS Is a reasonably INVOLVED Technique OF EQUATIONS WHICH WE'LL SOLVE Employing SUBSTITUTION.
SO I Made a decision to Fix The very first EQUATION HERE FOR Z.
SO I Extra THIS Expression TO Either side On the EQUATION And afterwards DIVIDED BY 0.
04 GIVING US THIS Price Listed here FOR Z BUT IF We discover THIS QUOTIENT AND Transfer Y Into the -two For the DENOMINATOR WE CAN ALSO Generate Z AS THIS Portion Right here.
NOW THAT WE KNOW Z IS EQUAL TO THIS Portion, We will SUBSTITUTE THIS FOR Z INTO THE SECOND EQUATION Below.
WHICH IS WHAT WE SEE In this article BUT Recognize HOW This really is RAISED To your EXPONENT OF -two SO THIS WOULD BE one, 225 For the -two DIVIDED BY Y Into the -4.
SO WE Might take THE RECIPROCAL WHICH WOULD GIVE US Y On the 4th DIVIDED BY one, five hundred, 625 AND HERE'S THE 21.
Given that Now we have AN EQUATION WITH JUST ONE VARIABLE Y WE WANT TO SOLVE THIS FOR Y.
SO FOR THE FIRST STEP, You will find there's Frequent Aspect OF Y.
SO Y = 0 WOULD SATISFY THIS EQUATION AND Could well be A Significant Stage BUT We all know We are NOT Heading TO HAVE A DIMENSION OF ZERO SO WE'LL JUST Disregard THAT Benefit AND SET THIS EXPRESSION Below Equivalent TO ZERO AND Clear up WHICH IS WHAT WE SEE Right here.
SO We'll ISOLATE THE Y CUBED Expression After which you can Dice ROOT Each side From the EQUATION.
Therefore if WE Include THIS FRACTION TO Either side With the EQUATION And afterwards CHANGE THE ORDER In the EQUATION This really is WHAT WE Might have AND NOW FROM HERE TO ISOLATE Y CUBED WE Really have to MULTIPLY Through the RECIPROCAL OF THIS FRACTION Right here.
SO Detect HOW THE LEFT Aspect SIMPLIFIES JUST Y CUBED Which Item Here's Roughly THIS VALUE Listed here.
SO NOW TO SOLVE FOR Y We might CUBE ROOT BOTH SIDES On the EQUATION OR RAISE Each side On the EQUATION For the 1/three Energy AND This offers Y IS About fourteen.
1918, AND NOW TO FIND THE Z COORDINATE In the CRITICAL POINT We could USE THIS EQUATION In this article Wherever Z = one, 225 DIVIDED BY Y SQUARED Which provides Z IS Roughly 6.
0822.
We do not Require IT Right this moment BUT I WENT Forward AND FOUND THE CORRESPONDING X Benefit At the same time USING OUR Quantity Method Clear up FOR X.
SO X WOULD BE About 4.
0548.
Since WE ONLY HAVE ONE Vital Place WE CAN In all probability Suppose THIS Level Will Limit The expense FUNCTION BUT TO Confirm THIS We are going to GO AHEAD AND Utilize the Important POINT AND The next Purchase OF PARTIAL DERIVATIVES JUST To ensure.
MEANING We are going to USE THIS System Below FOR D Along with the VALUES OF The 2nd Get OF PARTIAL DERIVATIVES TO DETERMINE No matter whether We've got A RELATIVE MAX OR MIN AT THIS Vital Issue WHEN Y IS Around fourteen.
19 AND Z IS About 6.
08.
Here i will discuss THE SECOND Purchase OF PARTIALS THAT WE Identified Before.
SO WE'LL BE SUBSTITUTING THIS Benefit FOR Y Which Benefit FOR Z INTO THE SECOND Purchase OF PARTIALS.
WE Ought to be Slightly Very careful THOUGH Simply because Try to remember WE HAVE A Perform 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 WOULD BE THE Z'S.
SO The next Get OF PARTIALS WITH Regard TO Y IS HERE.
The 2nd Purchase OF PARTIAL WITH RESPECT TO Z IS HERE.
HERE'S THE MIXED PARTIAL SQUARED.
Detect HOW IT Will come OUT To the POSITIVE Benefit.
Therefore if D IS Favourable AND SO IS THE SECOND PARTIAL WITH RESPECT TO Y Investigating OUR NOTES Listed here Which means We have now A RELATIVE Bare minimum AT OUR Essential Level And thus These are definitely THE DIMENSIONS That will Decrease The expense of OUR BOX.
THIS WAS THE X COORDINATE Within the Preceding SLIDE.
Here is THE Y COORDINATE AND This is THE Z COORDINATE WHICH Once again ARE THE DIMENSIONS OF OUR BOX.
Therefore the FRONT WIDTH Can be X WHICH IS Roughly four.
05 INCHES.
THE DEPTH Will be Y, That is APPROXIMATELY 14.
19 INCHES, AND THE HEIGHT Can be Z, Which can be APPROXIMATELY six.
08 INCHES.
LET'S FINISH BY Considering OUR Charge Purpose In which WE Contain the COST Perform With regard to Y AND Z.
IN A few DIMENSIONS This may BE THE Area Exactly where THESE Reduced AXES WOULD BE THE Y AND Z AXIS AND The expense Can be Alongside THE VERTICAL AXIS.
We can easily SEE THERE'S A Reduced POINT HERE AND THAT Transpired AT OUR Crucial POINT THAT WE Uncovered.
I HOPE YOU Observed THIS Handy.