- A RECTANGLE IS INSCRIBED WITH ITS BASE ON THE X AXIS
AND ITS UPPER CORNERS ON THE PARABOLA Y = 4 - X SQUARED.
WE WANT TO KNOW WHAT ARE THE DIMENSIONS OF THE RECTANGLE
WITH THE GREATEST AREA.
NOTICE HOW THIS GREEN RECTANGLE
DOES FIT THE CONDITIONS DESCRIBED.
SO LET'S CONSIDER THE COORDINATES OF THIS POINT
AND THIS POINT TO HELP US FIND THE DIMENSIONS OF THE RECTANGLE
IN TERMS OF X.
FOR THIS POINT WE COULD LABEL THE X COORDINATE X
AND THE Y COORDINATE,
BECAUSE IT'S ON THE FUNCTION, Y = 4 - X SQUARED.
THE Y COORDINATE MUST BE 4 - X SQUARED.
AND THEN FOR THE POINT ON THE LEFT,
NOTICE HOW THE Y COORDINATE WOULD BE THE SAME,
AND WE COULD LABEL THE X COORDINATE -X.
NOW, LET'S LIST THE LENGTH OF THE SIDES IN TERMS OF X.
WELL, THE Y COORDINATE WOULD BE THIS DISTANCE HERE,
WHICH WE KNOW MUST BE EQUAL TO 4 - X SQUARED.
AND THEN FOR THE LENGTH OF THESE TWO SIDES,
SINCE THE DISTANCE FROM THE Y AXIS TO THIS POINT WOULD BE X,
THE DISTANCE FROM THE Y AXIS TO THIS POINT ON THE LEFT
WOULD ALSO BE X.
AND THEREFORE, THE LENGTH OF THIS SIDE WOULD HAVE TO BE 2X.
THEREFORE, THE AREA OF THIS RECTANGLE IN TERMS OF X
WOULD BE EQUAL TO 2X x THE QUANTITY 4 - X SQUARED,
USING THE FORMULA LENGTH x WIDTH.
AND NOW, TO FIND THE DIMENSIONS THAT WILL MAXIMIZE THE AREA,
WE'LL USE CALCULUS TECHNIQUES,
WHICH MEANS WE'LL FIRST FIND THE CRITICAL NUMBERS,
WHICH IS WHERE THE DERIVATIVE WOULD BE UNDEFINED
OR EQUAL TO ZERO.
AND THEN USING THE CRITICAL NUMBERS,
WE'LL DETERMINE WHICH VALUES OF X WOULD MAXIMIZE THE AREA.
SO FOR THE NEXT STEP, WE'LL GO AHEAD AND CLEAR THE PARENTHESES.
SO WE'D HAVE "A" =--THIS WOULD BE 8X - 2X TO THE 3rd.
SO NOW, WE'LL FIND THE FIRST DERIVATIVE.
SO "A" PRIME = THE DERIVATIVE OF 8X,
WHICH WOULD BE 8 - THE DERIVATIVE OF 2X TO THE 3rd.
THAT WOULD BE 6X TO THE 2nd.
WELL, NOTICE HOW THIS IS NEVER UNDEFINED.
SO LET'S GO AHEAD AND SET IT EQUAL TO ZERO AND SOLVE FOR X.
LET'S GO AHEAD AND SUBTRACT 8 ON BOTH SIDES.
DIVIDE BOTH SIDES BY -6.
SO WE HAVE X SQUARED =--THIS SIMPLIFIES TO 4/3,
AND NOW, WE'LL TAKE THE SQUARE ROOT OF BOTH SIDES
TO SOLVE FOR X.
NOTICE X CAN BE BOTH POSITIVE OR NEGATIVE.
SO WE'D HAVE X = +/- --WELL, THE SQUARE ROOT OF 4 WOULD BE 2,
AND THEN WE'D HAVE THE SQUARE ROOT OF 3 IN THE DENOMINATOR.
NOTICE HOW WE HAVE TWO CRITICAL NUMBERS:
ONE IS POSITIVE SQUARE ROOT 2 DIVIDED BY 3,
AND ONE IS NEGATIVE SQUARE ROOT 2 DIVIDED BY 3.
BUT SINCE X REPRESENTS A LENGTH, WE KNOW X CAN'T BE NEGATIVE,
BUT LET'S GO AHEAD AND CONTINUE THIS ON THE NEXT SLIDE.
LET'S GO AHEAD AND RATIONALIZE THE DENOMINATOR.
SO WE'LL MULTIPLY BOTH THE TOP AND BOTTOM BY SQUARE ROOT 3.
SO AGAIN, WE HAVE TWO CRITICAL NUMBERS:
ONE IS X = -2 SQUARE ROOT 3/3,
AND THE OTHER IS X = +2 SQUARE ROOT 3/3.
SO AGAIN, BECAUSE WE KNOW X CAN'T BE NEGATIVE,
PROBABLY WOULD BE CORRECT TO ASSUME
THIS X VALUE WILL MAXIMIZE THE AREA.
BUT JUST TO BE SURE,
LET'S GO AHEAD AND USE THE SECOND DERIVATIVE TEST.
REMEMBER, OUR FIRST DERIVATIVE WAS EQUAL TO 8 - 6X SQUARED.
SO THE SECOND DERIVATIVE WOULD BE EQUAL TO--
THE DERIVATIVE OF 8 WOULD BE 0.
THE DERIVATIVE OF -6X SQUARED WOULD BE -12X.
SO NOTICE, "A" DOUBLE PRIME OF +2 SQUARE ROOT 3 DIVIDED BY 3
WOULD ACTUALLY BE NEGATIVE OR LESS THAN ZERO,
WHICH MEANS AT X = 2 SQUARE ROOT 3/3,
THE FUNCTION IS CONCAVE DOWN, WHICH IS GOOD NEWS,
BECAUSE THAT MEANS WE HAVE A MAXIMUM
WHEN X = 2 SQUARE ROOT 3 DIVIDED BY 3.
SO TO ANSWER THE QUESTION,
"WHAT ARE THE DIMENSIONS THAT WILL MAXIMIZE THE AREA,"
AGAIN, THE DIMENSIONS ARE 2X BY 4 - X SQUARED.
SO 2X WOULD BE EQUAL TO 2/1 x 2 SQUARE ROOT 3/3,
WHICH WOULD BE 4 SQUARE ROOT 3/3,
AND THEN 4 - X SQUARED
WOULD BE EQUAL TO 4 - 2 SQUARE ROOT 3/3 SQUARED.
SO WE'D HAVE 4 - --AND THEN 2 SQUARE ROOT 3 SQUARED.
THAT WOULD BE 4 x 3 OR 12.
3 SQUARED = 9.
SO THIS IS 12/9 OR 4/3, AND THEN 4 IS THE SAME AS 12/3.
12/3 - 4/3 = 8/3.
SO THE DIMENSIONS OF THE RECTANGLE WITH THE MAXIMUM AREA
WOULD BE 4 SQUARE ROOT 3 DIVIDED BY 3 UNITS BY 8/3 UNITS
OR FOR A DECIMAL APPROXIMATION,
DIMENSIONS WOULD BE APPROXIMATELY 2.3094 BY 2.6667.
THIS IS WHAT THE QUESTION ACTUALLY ASKED US TO FIND,
BUT BEFORE WE GO, LET'S GO AHEAD AND FIND
WHAT THAT MAXIMUM AREA WOULD BE.
AND SINCE THE AREA OF A RECTANGLE IS LENGTH x WIDTH,
THE MAXIMUM AREA WOULD BE EQUAL TO 4 SQUARE ROOT 3
DIVIDED BY 3 x 8/3.
NOTICE NOTHING SIMPLIFIES HERE.
SO THE AREA WOULD BE 32 SQUARE ROOT 3
DIVIDED BY 9 SQUARE UNITS.
OKAY. I HOPE YOU FOUND THIS EXPLANATION HELPFUL.