A printed poster is to have a total area of 711 square inches with top and bottom margins of 5 inches and side margins of 4 inches. What should be the dimensions of the poster so that the printed area be as large as possible? (Round your answers to three decimal places if necessary, but do not round until your final computation.) a. To solve, let ac denote the width of the poster in inches and let y denote the length in inches. We need to maximize the following function of :c and y. Area in terms of ac and y: b. We can reexpress this as the following function of m alone. f (73) : c. We nd that x) has a critical acvalue, denoted ace. 90c : d. To verify that f(ac) has a maximum at this critical number we compute the second derivative f\"(ac) and nd that its value at the critical number. are) : is a P v number e. We determine the optimal dimensions of the poster. ac : inches by y : inches f. Finally, this gives us a maximumal printed area. Area: square inches a. To solve, let :1: denote the width of the poster in inches and let y denote the length in inches. We need to maximize the following function of ac and y. b. We can reexpress this as the following function of m alone. c. We find that f (at) has a critical mvalue, denoted we. (EC: d. To verify that re) has a maximum at this critical number we compute the second derivative f\"(a;) and nd that its value at the critical number. f(a:c) = _ is a positive v 6) number e. We determine the optimal dimensions of the poster. Let the length of poster be y and width of poster be x . Total area of poster is my my = 711 sq inches ( eg- 1 ) ( given ) Area of pointed portion. It has topand bottom margin 5inches and side margins of 4 inches. Hence the pointed area is ( X - 2x 4 )x (y - 2x5) ( pointed area ) A .= 19-87 (4 - 10 ) We need to maximise Ap Ap = Hy - By-10x+ 80 = 711 - By -10 + 30 = 791 - By - 10x y = 711 from eg - 1 x Ap = 791 - 8x711- 10 2 for Ap to be maximum dAp = 0 dm 6 + 8 x711 - 10 = 0 x 2 = 8 x 71 1 : 7 X= 23 . 849 5 29 3 4: 711 10 y = 29. 8 11910 2 = 23. 85 0 inches y = 29. 312 inchesa) pointed pant area, is Ap = (7- 8 ) (4- 10) Ap = my - By- 107+ 80 b) +19) = (2- 07 ( 71 1-10) co y = =11 $ 1a) = 711 + 80- 10 x - 5688 x *(27 = 781 - 10 - 5608 x C ? XC = 23. 850 inches d ) + ( x ) = - 2 x 8 x 71 1 40 to x 2 70 X , 3 d" (21 ) = - 0 8385 4 1- 0. 8391 is a negative number e ) 1= 23. 850 inches y = 29. 812 inches + ! Ap = (2 - 87 ( 4 - 10 ) = 15 050 x 19. 012 = 314 0202 sq inong Ap = 314. 020 sq incho