Question

An animal breeder can buy four types of food for Vietnamese pot-bellied pigs. Each case of Brand A contains 25 units of fiber, 40 units of protein, and 40 units of fat. Each case of Brand B contains 50 units of fiber, 50 units of protein, and 30 units of fat. Each case of Brand C contains 75 units of fiber, 60 units of protein, and 30 units of fat. Each case of Brand D contains 100 units of fiber, 100 units of protein, and 50 units of fat. How many cases of each brand should the breeder mix together to obtain a food that provides 1275 units of fiber, 1140 units of protein, and 630 units of fat? Let x represent the number of cases of Brand A, y represent the number of cases of Brand B. z represent the number of cases of Brand C, and w represent be the number of cases of Brand D. There are four ways in which the breeder can mix brands to obtain a food that provides 1275 units of fiber, 1140 units of protein, and 630 units of fot. 1 w = 0, the solution is (0.0.0.0).

Answer 1

R2 R2-352 352 RI 252 25? 25 69.9 [os -14 -23.004 =) b= 23.004 = 1.643 Tu > 20 a=00046 Hence Y = (0.096) x4 + (1•643) * x=0 mph y = (0-046) (s) + (1.64317%) 19 = 197.15 feet) Hence answers are ra a=0.046) b= 16643 197.15 feet b. Stopping distance 69.9 25 625 a 1225 35 Ile

13 [ --6-9 - 9 2 3 os Since Columns > Row Rts Size Size 2*3] additive presse 6 9 -2 -3 8 A= 4 [1-3] B= (133) Ref ABZ [2][3] [49] + 2 Here AB is not Identity Hence given mabices are not inverse theter product = =[4]

AX=B [64] [%]=[3] Augomuted mabry [647] 6 R2 + R2-6R, [! -8 58 o -29 5 SO 2) X₂= -29 = -2 x + 642 = -8 = 2,24 Hence [-]