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Question Number 1155 by 123456 last updated on 05/Jul/15

what the min value of  60x+120y+1250z  if  x+y+5z=150  x∈{0,...,150}  y∈{0,...,150}  z∈{0,...,30}

$$\mathrm{what}\:\mathrm{the}\:\mathrm{min}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{60}{x}+\mathrm{120}{y}+\mathrm{1250}{z} \\ $$$$\mathrm{if} \\ $$$${x}+{y}+\mathrm{5}{z}=\mathrm{150} \\ $$$${x}\in\left\{\mathrm{0},...,\mathrm{150}\right\} \\ $$$${y}\in\left\{\mathrm{0},...,\mathrm{150}\right\} \\ $$$${z}\in\left\{\mathrm{0},...,\mathrm{30}\right\} \\ $$

Answered by prakash jain last updated on 08/Jul/15

min value z=0, y=0, x=150  60×150=9000

$$\mathrm{min}\:\mathrm{value}\:{z}=\mathrm{0},\:{y}=\mathrm{0},\:{x}=\mathrm{150} \\ $$$$\mathrm{60}×\mathrm{150}=\mathrm{9000} \\ $$

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