- YOU NEED TO MAXIMIZE THE AREA 
OF A CORRAL FOR YOUR PET.
HOWEVER, THERE IS A CLIFF 
ON ONE SIDE
SO YOU ONLY NEED TO FENCE 
THREE SIDES OF THE CORRAL.
SO IF THIS IS THE CORRAL,
LET'S ASSUME THIS RED DASH LINE 
HERE WOULD BE THE CLIFF,
SO YOU DON'T HAVE TO FENCE THIS 
SIDE OF THE RECTANGLE.
IF YOU HAVE 1,200 YARDS 
OF FENCING AVAILABLE,
WHAT ARE THE DIMENSIONS 
OF THE RECTANGLE
WITH THE MAXIMUM AREA 
AND WHAT IS THE MAXIMUM AREA?
BECAUSE WE DON'T KNOW THE LENGTH 
OF THE SIDES OF THE RECTANGLE,
LET'S GO AHEAD 
AND CALL THIS SIDE LENGTH X
AND BECAUSE THE OPPOSITE SIDES 
OF THE RECTANGLE
HAVE EQUAL LENGTH, 
THIS WOULD ALSO BE X.
AND NOW BECAUSE WE HAVE A TOTAL 
OF 1,200 YARDS OF FENCING,
THE LENGTH OF THIS SIDE HERE
WOULD HAVE TO BE THE TOTAL 
LENGTH - X - X OR 1,200 - 2X.
AGAIN THIS LENGTH HERE 
IS A TOTAL AMOUNT OF FENCING
MINUS THE LENGTH 
OF THE TWO SIDES HERE
WHICH IS REPRESENTED BY - 2X.
AND NOW BECAUSE 
THE AREA OF A RECTANGLE
IS EQUAL TO LENGTH TIMES WIDTH,
THEY CONSIDER THE AREA 
OF THIS RECTANGLE WOULD BE
"A" = X x THE QUANTITY 
1,200 - 2X.
LET'S GO AHEAD 
AND DISTRIBUTE HERE.
THAT WOULD GIVE US 
"A" = 1,200X - 2X SQUARED.
NOTICE HOW WE HAVE A QUADRATIC 
FUNCTION HERE,
SO LET'S WRITE THIS 
AS A FUNCTION.
WE'LL SAY "A" IS A FUNCTION 
OF X, OR A OF X EQUALS
NOW LET'S PUT THE TERMS 
IN DESCENDING ORDER
SO WE HAVE 
- 2X SQUARED + 1,200X.
AND NOW THAT OUR QUADRATIC 
IS IN THIS FORM HERE,
WE SHOULD BE ABLE TO RECOGNIZE
THAT THE COEFFICIENT OF X 
SQUARED OR "A" = - 2.
B, THE COEFFICIENT OF X = 1,200
AND BECAUSE 
THERE'S NO CONSTANT TERM
THAT MEANS C 
WOULD BE EQUAL TO 0.
SO BECAUSE WE HAVE A QUADRATIC 
FUNCTION,
AND "A" = - 2, 
OR A IS NEGATIVE,
WE SHOULD RECOGNIZE 
THAT THE GRAPH OF THIS QUADRATIC
WOULD OPEN DOWNWARD LIKE THIS
AND THEREFORE THE X COORDINATE 
OF THIS VERTEX HERE
WOULD ACTUALLY 
GIVE US THE LENGTH
OF THIS SIDE OF THE RECTANGLE 
THAT WOULD MAXIMIZE THE AREA
AND THE Y COORDINATE 
OF THE VERTEX
WOULD BE THAT MAXIMUM AREA.
SO LET'S GO AHEAD 
AND TAKE THIS FUNCTION HERE
UNDER THE NEXT SLIDE AND FIND 
THE COORDINATES OF THE VERTEX.
REMEMBER WHEN OUR QUADRATIC 
FUNCTION IS IN THIS FORM HERE,
THE X COORDINATE OF THE VERTEX 
IS -B DIVIDED BY 2A
AND THE Y COORDINATE 
OR THE FUNCTION VALUE
WOULD BE F OF -B DIVIDED BY 2A.
SO ONCE WE FIND THE X COORDINATE 
TO THE VERTEX,
WE'LL THEN SUB THAT VALUE 
INTO OUR FUNCTION
TO FIND THE Y COORDINATE 
OF THE VERTEX.
SO AGAIN WE ALREADY KNOW 
THAT "A" = -2 AND B = 1,200.
AND AGAIN OUR GOAL HERE
IS TO FIND THE COORDINATES 
OF THE VERTEX.
WE'LL BEGIN BY FINDING 
THE X COORDINATE OF THE VERTEX
SO WE'LL LET X EQUAL -B 
DIVIDED BY 2A WHERE B IS 1,200
SO WE HAVE -1,200 
DIVIDED BY 2 x "A" WHICH IS - 2.
SO THIS WOULD GIVE US 
-1,200 DIVIDED BY -4
WHICH = 300.
SO THE X COORDINATE
TO THE VERTEX IS 300,
SO IF WE GO BACK TO OUR SKETCH 
JUST FOR A MOMENT,
THAT MEANS THE LENGTH 
OF THIS SIDE
WOULD HAVE TO BE 300 YARDS
IN ORDER TO MAXIMIZE 
THE AREA OF THIS RECTANGLE.
NOW THAT WE KNOW X IS 300,
WE CAN ACTUALLY 
FIND THE LENGTH OF THIS SIDE
BY SUBBING 300 FOR X.
IF X IS 300, WE WOULD HAVE 
1,200 - 2 x 300 WHICH IS 600,
1,200 - 600 IS 600.
SO THIS LENGTH HERE 
WOULD HAVE TO BE 600 YARDS.
BUT WE'RE ALSO ASKED TO FIND 
THE MAXIMUM AREA
WHICH WOULD JUST BE THE PRODUCT 
OF 300 AND 600.
BUT IF WE GO BACK TO OUR WORK 
JUST FOR A MOMENT,
NOW THAT WE HAVE X = 300,
WE COULD ALSO FIND THE AREA 
BY FINDING THE Y COORDINATE
OR THE FUNCTION VALUE 
FOR FUNCTION "A" OF X.
LET'S JUST SHOW THAT WE WOULD 
GET THE SAME VALUE.
"A" OF 300 WOULD BE = -2 x 300 
SQUARED + 1,200 x 300.
SO THIS WOULD BE 300 SQUARED,
WHICH IS 90,000 x -2,
-180,000 + 1,200 x 300 
WOULD BE 360,000
WHICH IS EQUAL TO 180,000.
AND AGAIN JUST TO COMPARE, 
USING THE RECTANGLE,
WE KNOW THE AREA WOULD JUST BE 
300 x 600 SQUARE YARDS
AND OF COURSE 
300 x 600 IS 180,000.
AGAIN 
THIS WOULD BE SQUARE YARDS.
SO JUST TO SUMMARIZE,
THE DIMENSIONS 
THAT MAXIMIZE THE AREA
WOULD BE 300 YARDS BY 600 YARDS
AND THE MAXIMUM AREA 
IS 180,000 SQUARE YARDS.
LET'S GO AHEAD AND VERIFY 
THESE RESULTS GRAPHICALLY
BY GRAPHING OUR FUNCTION 
"A" OF X.
TO SAVE TIME 
I'VE ALREADY DONE THAT.
HERE IT IS.
WE CAN EITHER GRAPH IT IN 
FACTORED FORM OR EXPANDED FORM
BUT NOTICE HOW THE VERTEX HERE 
DOES HAVE AN X COORDINATE OF 300
WHICH REPRESENTS THE WIDTH 
THAT WOULD MAXIMIZE THE AREA
AND THEN 
THE Y COORDINATE 180,000,
REPRESENTS THE MAXIMUM AREA.
SO THIS DOES VERIFY OUR RESULTS.
OF COURSE IF WE WANTED TO 
WE COULD HAVE ALSO SOLVED THIS
USING A GRAPHING CALCULATOR
SO LET'S GO AHEAD 
AND QUICKLY SHOW THAT.
SO WE'LL PRESS Y EQUALS,
CLEAR ANY OLD FUNCTIONS 
AND TYPE IN OUR NEW FUNCTION
EITHER IN FACTORED FORM 
OR EXPANDED FORM.
I'M GOING TO TYPE 
IN 1,200X - 2X SQUARED,
AND WE DO HAVE TO CHANGE 
THE WINDOW.
TO SAVE TIME I'VE ALREADY 
ADJUSTED THE WINDOW.
NOTICE HOW THE X VALUES 
GO FROM -100 TO 700.
THE Y VALUES GO 
FROM -10,000 TO 200,000.
AND NOW IF WE PRESS GRAPH, WE 
CAN USE THE CALCULATION FEATURE
TO DETERMINE THE VERTEX.
WE CAN PRESS SECOND TRACE 
FOR CALCULATION
AND THEN WE WANT OPTION 4 
FOR MAXIMUM.
SO WE'LL PRESS 4 LEFT BOUND 
MEANS MOVE THE CURSOR
TO THE LEFT SIDE OF THE VERTEX 
SOMEWHERE HERE, PRESS ENTER,
MOVE TO THE RIGHT OF THE VERTEX, 
SOMEWHERE HERE, PRESS ENTER,
THEN ENTER ONE MORE TIME.
NOTICE HOW THAT CALCULATOR 
IS NOT PERFECT,
WE NEED TO RECOGNIZE 
THAT THIS DOES REPRESENT
AN X VALUE OF 300 
AND THE Y VALUE IS 180,000.
I HOPE YOU FOUND 
THIS EXPLANATION HELPFUL.
