Ex - Quadratic function application - Maximum area

- 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.