For solution-based tasks such as this one, only an image of a HANDWRITTEN output containing your solution and answers will be credited. Either scan or take a photo of your output. Do not submit your output that is encoded in any text editor. word processor such as Microsoft Word. If your output is not an image of a handwritten document, you will get ZERO points for it. Solve each of the ff. problems. Show your solution or else no points will be given to that specic item. Show a thorough solution similar to the ones shown in our class or in the presentation les or Excel les. For computed values that are decimal numbers, round off to 2 decimal places unless otherwise specied. (50 points each ) Attach the Honor Pledge for Graded Assignment. 1.) A food manufacturing plant orders up to one hundred kilograms of sh and two hundred forty kilograms of chicken per day for use in manufacturing its two types, Regular and Special. Production of one pack of the Regular type needs one kilogram of sh and three kilograms of chicken, while one pack of the special type requires two kilograms of sh and two kilograms of chicken. In view of the company's heavy investment in advertising of the regular type. it has been decided that at least one-third of the total production should be the Regular type. On the other hand, because of the consumer climate, the Special type is earning a prot of three hundred forty-ve pesos per pack, while the prot of the Regular type is one hundred seventy-eight pesos per pack. Given these restrictions, how many packs of each kind of variety should be produced to maximize prots, and what is the maximum possible prot? 2.) Thomas needs at least forty-eight units of protein, sixty units of carbohydrates, and fty units of fat each month. From each kilogram of Food A, he receives two units of protein, four units of carbohydrates, and ve units of fats. Food B contains three units of protein, three units of carbohydrates, and two units of fats. If Food A costs one hundred ten pesos per kilogram and Food B costs ninety pesos per kilogram, how many kilograms of each food should Thomas buy each month to keep costs at a minimum? What is this minimum cost? A local boutique produces gowns of two designs, A and B. The available materials for these are the ff: 18 square meters of cotton, 20 square meters of silk, and 5 square meters of wool. Each gown of Design A requires 3 square meters of cotton, 2 square meters of silk, and 1 square meter of wool. Each gown of Design B requires 2 square meters of cotton and 4 square meters of silk. Each gown of Design A sells for P1,200 while Design B for P1,600. If there is an advanced order of 2 gowns of Design A, how many gowns of each design should the boutique produce to obtain the maxintim arounE oto profile sign A gowns produced y = no. of Design B gowns produced Maximize P= 1200x+ 1600y Structural Constraints : P = profit Cotton: 3x +2y = 18 - Inequality 0 silk : 2x + 49 = 20 Materials Design A (X ) Design B (y ) Available * ( 2 x + Ey) =( 20) 2 Cotton 2 * + 29 = 10 - Ing. 3 18 NW Silk 4 (1 ) x + 09 5 5 20 Wool 5 X 5 5 - Ing () ( 24X =5 Advanced Order : X 2 2 - Ing @ Profit 1200 1600 of Design A gowns Non - negativity constraints: X 20, yzoTo graph 3 x + 2 y = 18, graph 3x+ 29 = 18 3 * + 29 = 18 Y when y = 0: 3 x + 2 (0 ) = 18 3 x+ 0 = 18 3 x = 18 X = 6 ( 6, 0 ) when x = 0 : 3 ( 0 ) + 29 = 18 Region 0+ 29= 18 of solution Using y = 9 ( 0, 9 ) X Test Point ( 0, 0), evaluate 3 x + 29 = 18 1 2/3 4 5 6 7 8 9 10 11 3 (0 ) + 2 10) = 18 0 = 18 True: So the Region of Solu for 3 x+ 29 4 18 includes (0,0)To graph x7 29 = 10, sketch x+ 24 = 10 ifyzo ; x+ 29 = 10 X + 2 (0) = 10 xto = 10 x = 10 ( 10, 0) jut if x = 0; of 29 = 10 24 = 10 4 = 5 ( 0, 5 ) Lotsolution Test Mount ( 0, 0) X 1 2 3 4 5 6 7 8 9 1071 729=10 X+ 24 = 10 X+ 24= 10 0+ 2 ( 0 ) = 10 Of TO True So, the region of solution for X+ 24 = 10 includes (0, 0)pt A : ( 21 0 ) pt B: intersection of x = 2 and X+ 24 = 10 substitute X= 2 in x+ 29= 10. 3 x + Zy = 18 2 + 29 = 10 9 29 = 10-2 8 29 = ] *+ 2yz/o 4= 4 A (2, 4 ) B Feasible HNWA UT at D: intersection of 3 x+ 24= 18 and X+ 2y= 10 Region 3 x+ 24 = 18 "7 subbt, x= $ in - x+ 29 = 10 * + 29 = 10 X 3 x- X + 24-29= 18-10 4+ 29 = 10 1 2 3 4 5 6 2 8 9 10 11 2 x + 0 y = 8 2 9 = 10 - 4 pt E: substitute x = 5 in 3x+ 2y= 18 *+ 29= 10 2x = 8 2 4 =6 3 (5) + 24=18 X= 4 y - 3 15+ 29= 18 ot F ( 5, 0 ) D ( 4, 3 ) 29= 18-15 = 3 y = = = 1.5 E ( 5, 1. 5 )Extreme Points Objective Function Maximize P= 120OX + 1600y A (2, 0 ) P= 1200( 2) + 1600(0) = 2400 B ( 2 , 4 ) P = 1200 ( 2) + 1600 (4) = 8, 800 D ( 4, 3 ) P = 1200 ( 4) + 1600 ( 3 ) = 9, 600 E ( 5 , 1. 5 ) P = 1200 ( 5 ) + 1600 ( 1. 5 ) = 8 , 400 F ( 5 , 0 ) P = 1200 ( 5 ) + 1600 ( 0 ) = 6000 Since the largest value of Pis 8,800, then that is the optimal value. and the optimal solution is x= 4 and y= 3 Aus: 14 gowns of DesignA and 3 gowns of Design B