Question Number 80347 by jagoll last updated on 02/Feb/20
$${let}\:{x}\:{and}\:{y}\:{be}\:{positif}\:{real}\:{number} \\ $$$${such}\:{that}\:\mathrm{1}\leqslant{x}+{y}\leqslant\mathrm{9}\:{and} \\ $$$${x}\leqslant\mathrm{2}{y}\leqslant\mathrm{3}{x}.\:{what}\:{is}\:{the}\: \\ $$$${largest}\:{value}\:{of}\:\:\:\frac{\mathrm{9}−{y}}{\mathrm{9}−{x}} \\ $$$$ \\ $$
Commented by john santu last updated on 02/Feb/20
$${i}\:{think}\:{you}\:{can}\:{solve}\:{by}\: \\ $$$${linear}\:{programming} \\ $$
Commented by mr W last updated on 02/Feb/20
$${what}\:{if}\:{the}\:{question}\:{is} \\ $$$${what}\:{is}\:{the}\:{largest}\:{and}\:{smallest}\:{value} \\ $$$${of}\:{x}^{\mathrm{2}} +{xy}−\mathrm{2}{x}−\mathrm{3}{y}\:? \\ $$
Commented by jagoll last updated on 03/Feb/20
$${by}\:{non}\:{linear}\:{programming} \\ $$$${hahahaha} \\ $$
Commented by mr W last updated on 03/Feb/20
$${let}\:{k}={x}^{\mathrm{2}} +{xy}−\mathrm{2}{x}−\mathrm{3}{y} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{xy}−\mathrm{2}{x}−\mathrm{3}{y}−{k}=\mathrm{0} \\ $$$${this}\:{is}\:{the}\:{equation}\:{of}\:{a}\:{hyperbola}. \\ $$$${making}\:{this}\:{curve}\:{touch}\:{the}\:{area} \\ $$$${given},\:{we}\:{can}\:{find}\:{the}\:{minimum} \\ $$$${and}\:{maximum}\:{value}\:{of}\:{k}. \\ $$$${k}_{{min}} =−\frac{\mathrm{169}}{\mathrm{40}} \\ $$$${k}_{{max}} =\mathrm{33} \\ $$
Commented by jagoll last updated on 04/Feb/20
$${sir}\:{k}_{{min}\:} {at}\:{point}\:?\: \\ $$
Commented by mr W last updated on 04/Feb/20
$${k}_{{min}} \:{is}\:{when}\:{the}\:{curve}\:{just}\:{touches}\:{the} \\ $$$${topmost}\:{line}\:\mathrm{4}. \\ $$
Answered by mr W last updated on 02/Feb/20
$$\mathrm{1}\leqslant{x}+{y}\leqslant\mathrm{9} \\ $$$$\frac{{x}}{\mathrm{2}}\leqslant{y}\leqslant\frac{\mathrm{3}{x}}{\mathrm{2}} \\ $$$${the}\:{area}\:{of}\:{x},\:{y}\:{is}\:{defined}\:{by}\:\mathrm{4}\:{lines}: \\ $$$${line}\:\mathrm{1}:\:{x}+{y}=\mathrm{1} \\ $$$${line}\:\mathrm{2}:\:{x}+{y}=\mathrm{9} \\ $$$${line}\:\mathrm{3}:\:{y}=\frac{{x}}{\mathrm{2}} \\ $$$${line}\:\mathrm{4}:\:{y}=\frac{\mathrm{3}{x}}{\mathrm{2}} \\ $$$$ \\ $$$${let}\:\frac{\mathrm{9}−{y}}{\mathrm{9}−{x}}=\frac{{y}−\mathrm{9}}{{x}−\mathrm{9}}={k} \\ $$$$\Rightarrow{y}=\mathrm{9}+{k}\left({x}−\mathrm{9}\right) \\ $$$${k}\:{is}\:{the}\:{inclination}\:{of}\:{a}\:{line}\:{passing} \\ $$$${the}\:{point}\:{S}\left(\mathrm{9},\mathrm{9}\right)\:{and}\:{the}\:{area}\:{described} \\ $$$${above}. \\ $$$${we}\:{see}\:{k}\:{is}\:{maximum}\:{when}\:{this}\:{line}\:\left({A}\right) \\ $$$${passes}\:{the}\:{point}\:{P}\:{and}\:{k}\:{is}\:{minimum} \\ $$$${when}\:{this}\:{line}\:\left({B}\right)\:{passes}\:{the}\:{point}\:{Q}. \\ $$$$ \\ $$$${point}\:{P}={intersection}\:{of}\:{line}\:\mathrm{3}\:{and}\:\mathrm{2}: \\ $$$${y}=\frac{{x}}{\mathrm{2}} \\ $$$${x}+{y}=\mathrm{9} \\ $$$$\Rightarrow{x}=\mathrm{6},\:{y}=\mathrm{3}\:\Rightarrow{P}\left(\mathrm{6},\mathrm{3}\right) \\ $$$${k}_{{max}} =\frac{\mathrm{9}−\mathrm{3}}{\mathrm{9}−\mathrm{6}}=\mathrm{2} \\ $$$${point}\:{Q}={intersection}\:{of}\:{line}\:\mathrm{4}\:{and}\:\mathrm{2}: \\ $$$${y}=\frac{\mathrm{3}{x}}{\mathrm{2}} \\ $$$${x}+{y}=\mathrm{9} \\ $$$$\Rightarrow{x}=\frac{\mathrm{18}}{\mathrm{5}},\:{y}=\frac{\mathrm{27}}{\mathrm{5}}\:\Rightarrow{Q}\left(\frac{\mathrm{18}}{\mathrm{5}},\frac{\mathrm{27}}{\mathrm{5}}\right) \\ $$$${k}_{{min}} =\frac{\mathrm{9}−\frac{\mathrm{27}}{\mathrm{5}}}{\mathrm{9}−\frac{\mathrm{18}}{\mathrm{5}}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$ \\ $$$$\Rightarrow\frac{\mathrm{2}}{\mathrm{3}}\leqslant\frac{\mathrm{9}−{y}}{\mathrm{9}−{x}}\leqslant\mathrm{2} \\ $$
Commented by mr W last updated on 02/Feb/20
Commented by jagoll last updated on 02/Feb/20
$${thank}\:{you}\:{mister} \\ $$