- A RECTANGULAR FIELD 
NEEDS TO BE ENCLOSED.
FENCING COSTS $4.00 PER FOOT 
FOR TWO OPPOSITE SIDES
AND $5.00 PER FOOT 
FOR THE OTHER TWO SIDES.
WE WANT TO FIND THE DIMENSIONS 
OF THE FIELD
WITH AN AREA OF 800 SQUARE FEET 
THAT WILL MINIMIZE THE COST.
LET'S SKETCH A MODEL 
FOR THIS SITUATION.
FIRST, WE KNOW WE HAVE 
A RECTANGLE
AND THEREFORE THE OPPOSITE SIDES 
ARE THE SAME LENGTH.
LET'S LET THESE TWO SIDES 
EQUAL X FEET,
LET THESE TWO SIDES 
EQUAL Y FEET.
WE KNOW THE AREA OF THIS FIELD 
MUST BE 800 SQUARE FEET.
THE AREA OF THIS RECTANGLE 
WOULD BE X x Y
WHICH AGAIN MUST EQUAL 
800 SQUARE FEET.
SO OUR CONSTRAINT 
FOR THIS PROBLEM IS X x Y OR XY
MUST EQUAL 800 SQUARE FEET.
AGAIN, THIS IS OFTEN CALLED 
THE CONSTRAINT.
NOW LETS TALK ABOUT THE COST.
NOW LET'S CONSIDER 
THE COST OF THE FENCE.
IT SAYS IT COSTS $4.00 PER FOOT 
FOR TWO OPPOSITE SIDES
AND $5.00 PER FOOT 
FOR THE OTHER TWO SIDES.
THIS OFTEN OCCURS IF THEY FACTOR 
IN THE COSTS OF GATES
FOR TWO SIDES OF THE ENCLOSURE.
SO IT DOESN'T MATTER 
HOW WE LABEL THIS.
LET'S GO AHEAD AND JUST ASSUME 
THAT THE SHORTER SIDES HERE
COST $5.00 PER FOOT
AND THE LONGER SIDES COST 
$4.00 PER FOOT.
NOW LET'S WRITE A COST EQUATION.
THE TOTAL COST OF THE ENCLOSURE
IS GOING TO BE EQUAL TO THE COST 
PER FOOT x THE NUMBER OF FEET.
SO IF WE FIRST CONSIDER 
THE COST OF THESE TWO SIDES
THE COST OF THIS FENCE HERE 
IS GOING TO BE X FEET
x $5.00 PER FOOT
WHICH WOULD JUST BE X x 5 OR 5X.
THE OPPOSITE SIDE 
IS GOING TO COST THE SAME 5X.
THEN THE REMAINING TWO SIDES 
OR THESE TWO SIDES HERE
ARE GOING TO COST 
$4.00 PER FOOT.
SO THE COST OF ONE LENGTH HERE 
IS GOING TO BE Y FEET
x $4.00 PER FOOT
WHICH WOULD JUST BE 4Y 
FOR THIS LENGTH
AND THEN ALSO 4Y 
FOR THIS LENGTH HERE.
SO WE HAVE + 4Y + 4Y.
SO OUR COST FUNCTION IS GOING 
TO BE EQUAL TO 10X + 8Y.
NOW THIS IS THE EQUATION 
THAT WE WANT TO MINIMIZE
BUT RIGHT NOW WE HAVE THE COST 
FUNCTION
IN TERMS OF BOTH X AND Y.
WE WANT TO BE ABLE TO REWRITE 
THIS IN TERMS OF ONE VARIABLE
USING A SUBSTITUTION 
FROM THE CONSTRAINT.
AGAIN, WE WANT TO MINIMIZE THIS.
THIS IS OFTEN CALLED 
THE PRIMARY EQUATION.
SO WE'LL HAVE TO GO BACK UP 
TO THE CONSTRAINT NOW
AND SOLVE IT FOR X OR Y.
LET'S GO AHEAD AND SOLVE THIS 
FOR Y
BY DIVIDING BOTH SIDES BY X.
SO WE WOULD HAVE Y = 
800 DIVIDED BY X
WHICH MEANS NOW WE CAN REPLACE 
THIS Y HERE
WITH 800 DIVIDED BY X
AND THEN WE'LL HAVE THE COST 
FUNCTION
IN TERMS OF ONE VARIABLE.
SO WE'LL HAVE THE COST 
= 10X + 8 x 800 DIVIDED BY X
AND NOW IN ORDER TO MINIMIZE 
THIS
WE WANT TO FIRST FIND 
THE CRITICAL NUMBERS
BY FINDING THE DERIVATIVE 
OF THIS FUNCTION
AND THEN DETERMINING 
WHERE IT'S EQUAL TO ZERO
OR WHERE IT'S UNDEFINED.
BUT BEFORE WE DO THAT LET'S GO 
AHEAD AND REWRITE THIS
AS C = 10X.
THIS IS LIKE 8/1 
SO THIS WOULD BE 6,400/X
OR + 6,400 X TO THE POWER OF -1.
IF WE WERE TO MOVE THIS X 
UP INTO THE NUMERATOR,
REMEMBER IT'S GOING TO CHANGE 
THE SIGN OF THE EXPONENT.
NOW LET'S GO AHEAD 
AND FIND THE DERIVATIVE OF THIS.
SO C PRIME IS GOING TO BE EQUAL 
TO THE DERIVATIVE OF 10X IS 10.
HERE WE'LL APPLY THE POWER RULE 
SO WE'LL MULTIPLY BY -1.
SO WE'LL HAVE - 6,400 AND THEN 
X TO THE POWER OF -1 - 1
WHICH WOULD BE X 
TO THE POWER OF -2.
WHICH IF WE WANTED TO 
WE COULD WRITE THIS AS
C PRIME = 10 - 6,400/X SQUARED.
SO THIS WOULD BE UNDEFINED 
WHEN X IS EQUAL TO ZERO
BUT IF X IS EQUAL TO ZERO
THE AREA OF THIS RECTANGLE 
WOULD BE ZERO
AND THEREFORE 
WE'RE NOT GOING TO CONSIDER THAT
AS A CRITICAL NUMBER.
SO NOW WE'RE GOING TO SET THIS 
EQUAL TO ZERO AND SOLVE FOR X.
LET'S GO AHEAD AND DO THAT 
ON THE NEXT SLIDE.
SO IF WE SET THIS EQUAL TO ZERO
LET'S GO AHEAD 
AND ADD THIS FRACTION
TO THE OTHER SIDE 
OF THE EQUATION
AND WE'LL WRITE 10 AS 10/1.
SO WE'D HAVE 10/1 EQUALS 6,400 
DIVIDED BY X SQUARED
AND NOW THAT WE HAVE 
TWO FRACTIONS
EQUAL TO EACH OTHER
THIS IS A PROPORTION
SO WE CAN CROSS MULTIPLY 
AND THEN SOLVE FOR X.
SO 10X SQUARED MUST EQUAL 6,400.
GO AND DIVIDE BOTH SIDES BY 10 
SO WE HAVE X SQUARED = 640
AND NOW WE'LL SQUARE ROOT 
BOTH SIDES OF THE EQUATION.
REMEMBER WE ONLY HAVE 
TO CONSIDER
THE PRINCIPLE SQUARE ROOT 
OR THE +SQUARE ROOT
SINCE REMEMBER X REPRESENTS 
A LENGTH
AND WE WANT TO ROUND THIS 
TO THE NEAREST 10th.
SO X IS APPROXIMATELY 25.3 FEET.
NOW WE NEED TO KEEP IN MIND
THAT WE ARE TRYING TO MINIMIZE 
THE COST
AND WE ONLY HAD ONE CRITICAL 
NUMBER TO CONSIDER
AND WE OFTEN ASSUME
THIS IS GOING TO MINIMIZE 
THE COST FUNCTION.
BUT REMEMBER THERE ARE A COUPLE 
OF WAYS TO MAKE SURE
THAT AT THIS X VALUE 
THE COST WOULD BE MINIMIZED
RATHER THAN MAXIMIZED.
ONE WAY IS TO CHECK THE SIGN 
OF THE SECOND DERIVATIVE
AT THIS X VALUE.
SO LET'S GO AHEAD 
AND TAKE A MOMENT AND DO THAT.
SO HERE'S OUR FIRST DERIVATIVE
WHICH WE'LL GO AHEAD AND REWRITE 
AS 10 - 6,400 X TO THE -2
AND WE'LL FIND 
THE SECOND DERIVATIVE.
AGAIN, JUST TO ENSURE THAT 
AT THAT X VALUE
WE DO HAVE A MINIMUM COST.
WELL THE DERIVATIVE OF 10 
WOULD BE ZERO.
APPLYING THE POWER RULE HERE 
WE WOULD MULTIPLY BY -2.
ITS GOING TO BE +12,800 
X TO THE -3
WHICH WE CAN NOW WRITE AS 12,800 
DIVIDED BY X TO THE 3rd.
SO AT C DOUBLE PRIME 
OF APPROXIMATELY 25.3
NOTICE HOW WE WOULD HAVE 12,800 
DIVIDED BY 25.3 CUBED
AND WE ONLY CARE ABOUT THE SIGN 
HERE.
SO THIS IS GOING TO BE POSITIVE 
OR GREATER THAN ZERO.
REMEMBER THAT TELLS US THE 
CONCAVITY OF THE COST FUNCTION.
IF THE SECOND DERIVATIVE 
IS POSITIVE
THAT MEANS THE FUNCTION WOULD BE 
CONCAVE UP
AND THEREFORE AT THE CRITICAL 
NUMBER 25.3 HERE
WE WOULD HAVE 
A MINIMUM FUNCTION VALUE.
THIS JUST ASSURES US THAT AT X = 
25.3 WE HAVE MINIMIZED THE COST.
BUT REMEMBER GOING BACK 
TO OUR ORIGINAL SKETCH
WE STILL HAVE TO DETERMINE 
THE VALUE OF Y
AND WE CAN DO THAT HERE
SINCE WE KNOW THAT Y 
IS EQUAL TO 800 DIVIDED BY X.
SO IF WE SUBSTITUTE 25.3 FOR X
WE WOULD HAVE 800 
DIVIDED BY 25.3.
WE'D HAVE Y IS APPROXIMATELY 
EQUAL TO THIS QUOTIENT
WHICH COMES OUT TO APPROXIMATELY 
31.6 FEET.
SO AGAIN NOW WE KNOW THAT 
X IS APPROXIMATELY 25.3 FEET
AND Y IS APPROXIMATELY 
31.6 FEET.
IF WE TAKE A LOOK 
AT OUR SKETCH
NOTICE HOW WE HAVE LESS FEET 
AT THE COST OF $5.00 PER FOOT
AND WE HAVE MORE FEET 
AT THE $4.00 PER FOOT COST.
SO LET'S GO AHEAD 
AND SUMMARIZE ALL THIS.
THE FIRST QUESTION WAS 
FIND THE DIMENSIONS OF THE FIELD
WITH AN AREA OF 800 
THAT WILL MINIMIZE THE COST.
WELL THE DIMENSIONS OF THE FIELD 
ARE X FEET BY Y FEET
OR 25.3 FEET BY 31.6 FEET.
NOW IT DOESN'T ASK
BUT LET'S GO AHEAD AND FIND 
WHAT THAT TOTAL COST WOULD BE.
REMEMBER OUR COST FUNCTION 
WAS EQUAL TO 10X + 8Y.
SO WE'D HAVE (10 x 25.3) 
+ (8 x 31.6)
WHICH COMES OUT TO $505.80.
OKAY. THAT'S GOING TO DO IT 
FOR THIS PROBLEM.
I HOPE YOU FOUND THIS HELPFUL.
