Question Number 209123 by Spillover last updated on 02/Jul/24
Answered by A5T last updated on 02/Jul/24
$${max}\left(\mathrm{1000}{x}+\mathrm{600}{c}\right)\:{given}\:{c}\geqslant\mathrm{4}{x}\:{and}\:{x}+{c}\leqslant\mathrm{200} \\ $$$$\mathrm{4}{x}−{c}\leqslant\mathrm{0}\:{and}\:{x}+{c}\leqslant\mathrm{200}\Rightarrow{x}\leqslant\mathrm{40} \\ $$$$\Rightarrow{max}\left(\mathrm{1000}{x}+\mathrm{600}{c}\right)=\mathrm{1000}×\mathrm{40}+\mathrm{600}×\mathrm{160} \\ $$$$=\mathrm{136000}. \\ $$
Commented by Spillover last updated on 02/Jul/24
$${thank}\:{you}\: \\ $$
Answered by Spillover last updated on 02/Jul/24
