Answer to Question #169270 in Operations Research for gFg

Question #169270

10. A can of cat food, guaranteed by the manufacturer to contain at least 10 units of protein, 20 units of mineral matter, and 6 units of fat, consists of a mixture of four different ingredients. Ingredient A contains 10 units of protein, 2 units of mineral matter, and 1 2 unit of fat per 100g. Ingredient B contains 1 unit of protein, 40 units of mineral matter, and 3 units of fat per 100g. Ingredient C contains 1 unit of protein, 1 unit of mineral matter, and 6 units of fat per 100g. Ingredient D contains 5 units of protein, 10 units of mineral matter, and 3 units of fat per 100g. The cost of each ingredient is Birr 3, Birr 2, Birr 1, and Birr 4 per 100g, respectively. How many grams of each should be used to minimize the cost of the cat food, while still meeting the guaranteed composition? (Hint: Solve through simplex model)


1
Expert's answer
2021-03-09T17:06:52-0500

"10A+B+C+5D\u226510,"

"2A+40B+C+10D\u226520,"

"12A+3B+6C+3D\u22656,"

"F = 3A+2B+C+4D~\\rightarrow min,"

"\\begin{cases}\n -10A-B-C-5D+K=-10\\\\\n -2A-40B-C-10D+L=-20\\\\\n-12A-3B-6C-3D+M=-6\n\\end{cases}" "\\def \\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c:c:c:c:c}\n C& 3 & 2&1&4&0&0&0&0 \\\\ \\hline\n bas& A & B&C&D&K&L&M&b\\\\\n \\hline\n K & -10 & -1&-1&-5&1&0&0&-10\\\\ \\hline\nL&-2&-40&-1&-10&0&1&0&-20\\\\ \\hline\nM&-12&-3&-6&-3&0&0&1&-6\\\\\\hline\n\\end{array}" "|b|_{max}=|-20|"


"\\def \\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c:c:c:c:c}\n C& 3 & 2&1&4&0&0&0&0 \\\\ \\hline\n bas& A & B&C&D&K&L&M&b\\\\\n \\hline\n K & -\\frac{199}{20}& 0&-\\frac{39}{40}&-\\frac{19}{4}&1&-\\frac{1}{40}&0&-\\frac{19}{2}\\\\ \\hline\nB&\\frac{1}{20}&1&\\frac{1}{40}&\\frac 14&0&-\\frac{1}{40}&0&\\frac 12\\\\ \\hline\nM&-\\frac{237}{20}&0&-\\frac{237}{40}&-\\frac 94&0&-\\frac{3}{40}&1&-\\frac 92\\\\\\hline\n\\end{array}" "|b|_{max}=|-\\frac{19}{2}|"


"\\def \\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c:c:c:c:c}\n C& 3 & 2&1&4&0&0&0&0 \\\\ \\hline\n bas& A & B&C&D&K&L&M&b\\\\\n \\hline\n A & 1& 0&\\frac{39}{398}&\\frac{95}{199}&-\\frac{20}{199}&\\frac{1}{398}&0&\\frac{190}{199}\\\\ \\hline\nB&0&1&\\frac{4}{199}&\\frac {45}{199}&\\frac{1}{199}&-\\frac{5}{199}&0&\\frac {90}{199}\\\\ \\hline\nM&0&0&-\\frac{948}{199}&\\frac {678}{199}&-\\frac{237}{199}&-\\frac{9}{199}&1&\\frac {1356}{199}\\\\\\hline\n\\end{array}"

"\\def \\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c:c:c:c:c}\n C& 3 & 2&1&4&0&0&0&0 \\\\ \\hline\n bas& A & B&C&D&K&L&M&b\\\\\n \\hline\n A & 1& 0&\\frac{39}{398}&\\frac{95}{199}&-\\frac{20}{199}&\\frac{1}{398}&0&\\frac{190}{199}\\\\ \\hline\nB&0&1&\\frac{4}{199}&\\frac {45}{199}&\\frac{1}{199}&-\\frac{5}{199}&0&\\frac {90}{199}\\\\ \\hline\nM&0&0&-\\frac{948}{199}&\\frac {678}{199}&-\\frac{237}{199}&-\\frac{9}{199}&1&\\frac {1356}{199}\\\\\\hline\n\\Delta&0&0&-\\frac{265}{398}&-\\frac {421}{199}&-\\frac{58}{199}&-\\frac{17}{398}&0&\\frac {750}{199}\\\\\\hline\n\\end{array}"


"A=\\frac{190}{199}, B =\\frac{ 90}{199}, C = 0, D = 0,"

"F=3\\cdot \\frac{190}{199}+2\\cdot \\frac{90}{199}+1\\cdot 0+4\\cdot 0=3\\frac{153}{199}\\approx3.7688~\\text{Birr}."


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