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- A NORMAN WINDOW IS CONSTRUCTED BY ADJOINING A SEMICIRCLE
TO THE TOP OF AN ORDINARY RECTANGULAR WINDOW
AS WE SEE PICTURED HERE.
WE WANT TO FIND THE DIMENSIONS OF A NORMAN WINDOW
OF MAXIMUM AREA
IF THE TOTAL PERIMETER IS 16 FEET.
SO TO SET THIS UP
WE'RE GOING TO LET THE DIMENSIONS OF THE RECTANGLE
BE L FEET BY W FEET.
SO NOTICE IF THIS LENGTH IS W
THEN THE DIAMETER OF THE CIRCLE WOULD BE W
AND THE RADIUS WOULD BE 1/2W.
SO WE'LL FIRST SET UP THE CONSTRAINT FROM THIS PROBLEM
WHICH IS BASED UPON THE PERIMETER
WHICH MUST BE EQUAL TO 16 FEET.
PERIMETER WILL BE THE DISTANCE AROUND THE WINDOW
OR THIS DISTANCE HERE.
SO THREE SIDES WILL COME FROM THE RECTANGLE
AND THIS EDGE HERE WILL BE 1/2 THE CIRCUMFERENCE OF A CIRCLE
WITH DIAMETER OF W.
SO THE PERIMETER MUST BE EQUAL TO 2L + W +
AGAIN, HALF THE CIRCUMFERENCE OF A CIRCLE WITH DIAMETER W
AND SINCE THE CIRCUMFERENCE OF A CIRCLE IS EQUAL TO PI x DIAMETER
WE WOULD HAVE 1/2 PI x DIAMETER WHERE THE DIAMETER IS W,
AND THIS MUST BE EQUAL TO 16 FEET.
THIS IS OUR CONSTRAINT.
NOW OUR GOAL HERE IS TO MAXIMIZE THE AREA.
SO NOW WE'LL SET UP THE AREA EQUATION FOR THIS WINDOW.
THE AREA WOULD BE EQUAL TO THE AREA OF THE RECTANGLE
+ HALF THE AREA OF A CIRCLE WITH RADIUS 1/2W.
SO THE AREA OF THE RECTANGLE WOULD BE LENGTH x WIDTH OR LW
+ HALF THE AREA OF A CIRCLE WITH RADIUS 1/2
WHERE THE AREA OF A CIRCLE IS EQUAL TO PI R SQUARED.
SO WE'LL HAVE 1/2 PI R SQUARED OR 1/2 PI x R SQUARED
WHERE IS R IS 1/2W.
NOW TO MAXIMIZE THE AREA
WE WANT TO FIND THE DERIVATIVE OF THE AREA EQUATION
BUT BEFORE WE DO THIS,
WE NEED TO WRITE "A" IN TERMS OF ONE VARIABLE.
NOTICE RIGHT NOW IT'S IN TERMS OF L AND W.
SO WHAT WE'LL DO IS SOLVE THE CONSTRAINT FOR L
AND THEN PERFORM SUBSTITUTION FOR THIS L HERE.
SO TO SOLVE THE CONSTRAINT FOR L
NOTICE I'LL HAVE TO SUBTRACT THESE TWO W TERMS
TO THE RIGHT SIDE OF THE EQUATION
WHICH WOULD GIVE US 2L = 16 - W - 1/2 PI W
AND NOW TO SOLVE FOR L
WE'LL MULTIPLY BOTH SIDES OF THE EQUATION BY 1/2.
SO WE'LL HAVE L = 8 - 1/2W - THIS WOULD BE 1/4 PI W
AND NOW WE CAN SUBSTITUTE THIS FOR L INTO THE AREA EQUATION.
SO WE'LL HAVE THE AREA EQUALSTHE QUANTITY 8 - 1/2W - 1/4 PI W
x W + 1/2 SQUARED IS 1/4,
1/4 x 1/2 WOULD BE 1/8 SO + 1/8 PI W SQUARED.
SO NOW LETS GO AHEAD AND TAKE THIS EQUATION TO THE NEXT SLIDE,
SIMPLIFY AND THEN FIND THE FIRST DERIVATIVE
TO FIND THE CRITICAL NUMBERS OF THIS FUNCTION.
LET'S FIRST CLEAR THE PARENTHESES
BY DISTRIBUTING THE W.
SO WE'LL HAVE "A" = 8W - 1/2W SQUARED
MINUS 1/4 PI W SQUARED + 1/8 PI W SQUARED.
NOTICE HOW THESE LAST TWO TERMS ARE LIKE TERMS.
THIS IS EQUIVALENT TO -2/8 PI W SQUARED + 1/8 PI W SQUARED
SO THIS SIMPLIFIES TO - 1/8 PI W SQUARED
AND NOW TO FIND THE MAXIMUM VALUE OF "A"
WE'LL FIND THE FIRST DERIVATIVE
AND DETERMINE WHERE THE DERIVATIVE IS UNDEFINED
OR EQUAL TO ZERO.
SO "A" PRIME OR THE DERIVATIVE OF "A" WITH RESPECT TO W
IS EQUAL TO 8.
THIS WOULD BE - W AND HERE WE'D HAVE - 2/8 OR - 1/4 PI W.
WELL THIS IS NEVER GOING TO BE UNDEFINED FOR ANY VALUE OF W
SO NOW WE'LL SET THIS EQUAL TO ZERO AND SOLVE FOR W
TO FIND OUR CRITICAL NUMBER.
LET'S MOVE THE TWO W TERMS TO THE LEFT SIDE OF THE EQUATION
BY ADDING THEM TO BOTH SIDES.
SO THAT WOULD GIVE US W + 1/4 PI W = 8.
NOW LET'S FACTOR OUT THE COMMON FACTOR OF W.
SO WE HAVE W x THE QUANTITY 1 + 1/4 PI = 8
AND NOW WE'LL DIVIDE BOTH SIDES OF THE EQUATION BY THE QUANTITY
1 + 1/4 PI TO SOLVE FOR W.
SO W IS EQUAL TO THIS FRACTION HERE OR THIS QUOTIENT
BUT LET'S GO AHEAD AND CLEAR THE FRACTION FROM THE DENOMINATOR
BY MULTIPLYING THE NUMERATOR AND DENOMINATOR BY 4.
SO THE CRITICAL NUMBER IS W = 32 DIVIDED BY THE QUANTITY 4 + PI.
NOW THERE'S A COUPLE OF WAYS TO DETERMINE WHETHER THIS VALUE
PRODUCES A MAXIMUM OR MINIMUM AREA.
ONE WAY WOULD BE TO GRAPH THE AREA FUNCTION AND ANALYZE THIS.
WE SHOULD RECOGNIZE THIS IS A QUADRATIC
WHERE THE LEADING COEFFICIENT "A" WOULD BE NEGATIVE
AND THEREFORE IT'S A PARABOLA OPENING DOWN
AND THEREFORE WE WOULD HAVE A MAXIMUM VALUE AT THIS VALUE OF W
WHICH WE CAN SEE FROM THIS GRAPH HERE.
THE CRITICAL NUMBER IS HERE PRODUCING A MAXIMUM AREA.
BUT LET'S ALSO USE THE SECOND DERIVATIVE TEST
TO DETERMINE WHETHER WE HAVE A MAX OR MIN.
"A" DOUBLE PRIME
WOULD BE THE DERIVATIVE OF THE FIRST DERIVATIVE
WHICH WOULD BE 0 - 1 AND THEN THIS WOULD BE - 1/4 PI.
WELL THIS VALUE IS ALWAYS LESS THAN ZERO
WHICH MEANS THE FUNCTION IS ALWAYS CONCAVE DOWN
WHICH WE JUST SAW FROM THE GRAPH
AND IF IT'S CONCAVE DOWN
WE KNOW WE MUST HAVE A MAXIMUM VALUE AT THIS CRITICAL NUMBER.
THIS IS THE WIDTH OF THE RECTANGLE
THAT WILL PRODUCE THE MAXIMUM AREA FOR OUR NORMAN WINDOW,
AND THIS DOES COME OUT TO APPROXIMATELY 4.48,
WHICH WOULD BE FEET.
SO NOW THAT WE FOUND W
WE'LL GO BACK AND FIND L THE LENGTH OF THE RECTANGLE
GIVEN BY THIS EQUATION HERE
AND THEN WE'LL FIND THE RADIUS OF THE SEMICIRCLE
WHICH IS EQUAL TO 1/2W.
SO AGAIN, WE ALREADY KNOW W.
HERE'S THE EXACT VALUE AND THE APPROXIMATE VALUE IS 4.48 FEET
AND NOW WE'LL FIND THE RADIUS WHICH IS EQUAL TO 1/2W
OR 1/2 x 32/4 + PI.
THE EXACT VALUE WOULD BE 16 DIVIDED BY 4 + PI,
WHICH IS APPROXIMATELY 2.24 FEET.
WHICH SHOULD MAKE SENSE, THIS IS HALF OF W,
WHICH WE FOUND ABOVE,
AND NOW WE'LL FIND L USING OUR FORMULA HERE.
SO AGAIN, WE'LL SUBSTITUTE 32/4 + PI FOR W
AND BECAUSE WE'RE GOING TO COMBINE THESE
I'M GOING TO WRITE 8 AS 8/1.
SO WE'D HAVE 8/1 AND THEN - THIS WOULD BE 16/4 + PI
AND THIS WOULD BE 32 DIVIDED BY 4 IS EQUAL TO 8.
SO WE HAVE - 8 PI/4 + PI.
SO TO SUBTRACT THESE FRACTIONS
WE HAVE TO HAVE A COMMON DENOMINATOR
WHICH WOULD BE 4 + PI.
SO WE'LL MULTIPLY 8/1 BY 4 + PI/4 + PI.
NOW THAT WE HAVE A COMMON DENOMINATOR,
THIS WOULD BE 32 + 8 PI
AND THEN - 16 - 8 PI.
THE REASON I WANTED TO SHOW THIS
IS NOTICE HOW WE HAVE 8 PI - 8 PI THAT'S ZERO
AND 32 - 16 IS EQUAL TO 16.
SO L IS EQUAL TO 16 DIVIDED BY THE QUANTITY 4 + PI
WHICH IS ACTUALLY THE SAME AS R OR HALF OF W.
THIS IS GOING TO BE APPROXIMATELY 2.24 FEET
WHICH IS PROBABLY NOT AN EXPECTED OUTCOME.
SO LET'S GO AHEAD AND FINISH BY LABELING OUR NORMAN WINDOW
AND THEN CHECKING THE PERIMETER TO MAKE SURE IT'S 16 FEET
AND THEN DETERMINING THE MAXIMUM AREA.
SO AGAIN, W IS EQUAL TO APPROXIMATELY 4.48 FEET
AND HALF OF W IS APPROXIMATELY 2.24 FEET
AND WE FOUND THAT HALF OF W WAS ALSO EQUAL TO L.
SO L IS ALSO APPROXIMATELY 2.24 FEET
AND NOW TO CHECK THIS LET'S MAKE SURE THAT THE PERIMETER
WOULD BE 16 FEET
AND LET'S ALSO DETERMINE THE MAXIMUM AREA.
WELL HERE'S THE PERIMETER FORMULA
THAT WE HAD AS OUR CONSTRAINT
WHICH WE KNEW HAD BE EQUAL TO 16.
SO IF WE SUBSTITUTE L AND W INTO THIS EQUATION
NOTICE HOW EVEN THOUGH WE USED
THE APPROXIMATE VALUES OF L AND W
THE PERIMETER DOES COME OUT TO 16 FEET.
TO DETERMINE OUR MAXIMUM AREA
AGAIN, HERE'S THE AREA FORMULA FROM THE FIRST SLIDE.
WE'LL SUBSTITUTE L AND W INTO THIS EQUATION.
NOTICE IF WE DO THIS,
AGAIN WE'RE USING THE APPROXIMATE VALUES.
THE MAXIMUM AREA IS APPROXIMATELY 17.92 SQUARE FEET
WHICH AGAIN, IF WE GO BACK AND TAKE A LOOK AT THE GRAPH
OF THE AREA FUNCTION
ONCE YOU WROTE IT IN TERMS OF W HERE'S THE GRAPH
AND IT DOES LOOK LIKE THAT THE MAXIMUM AREA
IS JUST UNDER 18 SQUARE FEET.
OKAY. THAT'S GOING TO DO IT. I HOPE YOU FOUND THIS HELPFUL.