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Question Number 1151 by 123456 last updated on 05/Jul/15
what the min value of  60x+120y  where  x+y=150  x∈{0,1,...,150}  y∈{0,1,...,150}
$$\mathrm{what}\:\mathrm{the}\:\mathrm{min}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{60}{x}+\mathrm{120}{y} \\ $$$$\mathrm{where} \\ $$$${x}+{y}=\mathrm{150} \\ $$$${x}\in\left\{\mathrm{0},\mathrm{1},…,\mathrm{150}\right\} \\ $$$${y}\in\left\{\mathrm{0},\mathrm{1},…,\mathrm{150}\right\} \\ $$
Answered by prakash jain last updated on 05/Jul/15
y=150−x  60x+120×(150−x)  =120×150−60x  min value at x=150  =18000−9000=9000
$${y}=\mathrm{150}−{x} \\ $$$$\mathrm{60}{x}+\mathrm{120}×\left(\mathrm{150}−{x}\right) \\ $$$$=\mathrm{120}×\mathrm{150}−\mathrm{60}{x} \\ $$$$\mathrm{min}\:\mathrm{value}\:\mathrm{at}\:{x}=\mathrm{150} \\ $$$$=\mathrm{18000}−\mathrm{9000}=\mathrm{9000} \\ $$

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