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Question Number 79417 by mr W last updated on 25/Jan/20

For x,y∈R find the minimum and  maximum of 2x^2 −3x+4y  if x^2 +2y^2 −xy−5x−7y−30=0.

$${For}\:{x},{y}\in\mathbb{R}\:{find}\:{the}\:{minimum}\:{and} \\ $$$${maximum}\:{of}\:\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}{y} \\ $$$${if}\:{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} −{xy}−\mathrm{5}{x}−\mathrm{7}{y}−\mathrm{30}=\mathrm{0}. \\ $$

Commented by john santu last updated on 25/Jan/20

what the answer?

$${what}\:{the}\:{answer}? \\ $$

Commented by mr W last updated on 25/Jan/20

it′s not worked out yet. i have only  an idea, maybe not a good one.  basically i wanted to know which  methods we can apply to solve. what′s  your suggestion sir?

$${it}'{s}\:{not}\:{worked}\:{out}\:{yet}.\:{i}\:{have}\:{only} \\ $$$${an}\:{idea},\:{maybe}\:{not}\:{a}\:{good}\:{one}. \\ $$$${basically}\:{i}\:{wanted}\:{to}\:{know}\:{which} \\ $$$${methods}\:{we}\:{can}\:{apply}\:{to}\:{solve}.\:{what}'{s} \\ $$$${your}\:{suggestion}\:{sir}? \\ $$

Commented by john santu last updated on 25/Jan/20

yes sir, I also haven′t solved  it yet. with the Lagrange   multiplier method, i can′t yet

$${yes}\:{sir},\:{I}\:{also}\:{haven}'{t}\:{solved} \\ $$$${it}\:{yet}.\:{with}\:{the}\:{Lagrange}\: \\ $$$${multiplier}\:{method},\:{i}\:{can}'{t}\:{yet} \\ $$

Answered by mr W last updated on 26/Jan/20

=== General way to solve  ===  it is to find the minimum and  maximum of the function   F(x,y)=2x^2 −3x+4y  under the condition that  G(x,y)=x^2 +2y^2 −xy−5x−7y−30=0  Using Lagrange multiplier method  L(x,y)= 2x^2 −3x+4y+λ(x^2 +2y^2 −xy−5x−7y−30)  (∂L/∂x)=4x−3+λ(2x−y−5)=0    ...(i)  (∂L/∂y)=4+λ(4y−x−7)=0   ...(ii)  (∂L/∂λ)=x^2 +2y^2 −xy−5x−7y−30=0   ...(iii)  from (i) and (ii):  ((3−4x)/(2x−y−5))=(4/(x−4y+7))  3x−12y+21−4x^2 +16xy−28x=8x−4y−20  4x^2 −16xy+33x+8y−41=0   ...(iv)  x^2 +2y^2 −xy−5x−7y−30=0   ...(iii)  we get from (iii) and (iv) following  solution points (numerically):  (x,y)=(−3.5386, 1.6666) ⇒F=42.3256  (x,y)=(0.1194, 6.0763) ⇒F=23.9755  (x,y)=(11.3374, 4.8863) ⇒F=242.6063  (x,y)=(0.7960, −2.5756) ⇒F=−11.4232  ⇒f_(min) =−11.4232  ⇒f_(max) =242.6063

$$===\:{General}\:{way}\:{to}\:{solve}\:\:=== \\ $$$${it}\:{is}\:{to}\:{find}\:{the}\:{minimum}\:{and} \\ $$$${maximum}\:{of}\:{the}\:{function}\: \\ $$$${F}\left({x},{y}\right)=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}{y} \\ $$$${under}\:{the}\:{condition}\:{that} \\ $$$${G}\left({x},{y}\right)={x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} −{xy}−\mathrm{5}{x}−\mathrm{7}{y}−\mathrm{30}=\mathrm{0} \\ $$$${Using}\:{Lagrange}\:{multiplier}\:{method} \\ $$$${L}\left({x},{y}\right)=\:\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}{y}+\lambda\left({x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} −{xy}−\mathrm{5}{x}−\mathrm{7}{y}−\mathrm{30}\right) \\ $$$$\frac{\partial{L}}{\partial{x}}=\mathrm{4}{x}−\mathrm{3}+\lambda\left(\mathrm{2}{x}−{y}−\mathrm{5}\right)=\mathrm{0}\:\:\:\:...\left({i}\right) \\ $$$$\frac{\partial{L}}{\partial{y}}=\mathrm{4}+\lambda\left(\mathrm{4}{y}−{x}−\mathrm{7}\right)=\mathrm{0}\:\:\:...\left({ii}\right) \\ $$$$\frac{\partial{L}}{\partial\lambda}={x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} −{xy}−\mathrm{5}{x}−\mathrm{7}{y}−\mathrm{30}=\mathrm{0}\:\:\:...\left({iii}\right) \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right): \\ $$$$\frac{\mathrm{3}−\mathrm{4}{x}}{\mathrm{2}{x}−{y}−\mathrm{5}}=\frac{\mathrm{4}}{{x}−\mathrm{4}{y}+\mathrm{7}} \\ $$$$\mathrm{3}{x}−\mathrm{12}{y}+\mathrm{21}−\mathrm{4}{x}^{\mathrm{2}} +\mathrm{16}{xy}−\mathrm{28}{x}=\mathrm{8}{x}−\mathrm{4}{y}−\mathrm{20} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{16}{xy}+\mathrm{33}{x}+\mathrm{8}{y}−\mathrm{41}=\mathrm{0}\:\:\:...\left({iv}\right) \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} −{xy}−\mathrm{5}{x}−\mathrm{7}{y}−\mathrm{30}=\mathrm{0}\:\:\:...\left({iii}\right) \\ $$$${we}\:{get}\:{from}\:\left({iii}\right)\:{and}\:\left({iv}\right)\:{following} \\ $$$${solution}\:{points}\:\left({numerically}\right): \\ $$$$\left({x},{y}\right)=\left(−\mathrm{3}.\mathrm{5386},\:\mathrm{1}.\mathrm{6666}\right)\:\Rightarrow{F}=\mathrm{42}.\mathrm{3256} \\ $$$$\left({x},{y}\right)=\left(\mathrm{0}.\mathrm{1194},\:\mathrm{6}.\mathrm{0763}\right)\:\Rightarrow{F}=\mathrm{23}.\mathrm{9755} \\ $$$$\left({x},{y}\right)=\left(\mathrm{11}.\mathrm{3374},\:\mathrm{4}.\mathrm{8863}\right)\:\Rightarrow{F}=\mathrm{242}.\mathrm{6063} \\ $$$$\left({x},{y}\right)=\left(\mathrm{0}.\mathrm{7960},\:−\mathrm{2}.\mathrm{5756}\right)\:\Rightarrow{F}=−\mathrm{11}.\mathrm{4232} \\ $$$$\Rightarrow{f}_{{min}} =−\mathrm{11}.\mathrm{4232} \\ $$$$\Rightarrow{f}_{{max}} =\mathrm{242}.\mathrm{6063} \\ $$

Commented by john santu last updated on 25/Jan/20

i have done it like this, but it  was contrained by the numerical  method

$${i}\:{have}\:{done}\:{it}\:{like}\:{this},\:{but}\:{it} \\ $$$${was}\:{contrained}\:{by}\:{the}\:{numerical} \\ $$$${method} \\ $$

Answered by mr W last updated on 27/Jan/20

=== An other way to solve ===  x^2 −(y+5)x+2y^2 −7y−30=0  Δ=(y+5)^2 −4(2y^2 −7y−30)≥0  ⇒7y^2 −38y−145≤0  y_(1,2) =((19±4(√(86)))/7)  ⇒((19−4(√(86)))/7)≤y≤((19+4(√(86)))/7)  ⇒y_m =((19)/7)  2y^2 −(x+7)y+x^2 −5x−30=0  Δ=(x+7)^2 −8(x^2 −5x−30)≥0  ⇒7x^2 −54x−289≥0  x_(1,2) =((27±8(√(43)))/7)  ⇒((27−8(√(43)))/7)≤x≤((27+8(√(43)))/7)  ⇒x_m =((27)/7)  ⇒the points lie on an ellipse with  center at (((27)/7),((19)/7))  let x=u+((27)/7), y=v+((19)/7)  (u+((27)/7))^2 −(v+((54)/7))(u+((27)/7))+(2v−((11)/7))(v+((19)/7))−30=0  ⇒u^2 −uv+2v^2 −((344)/7)=0  let u=r cos ϕ, v=r sin ϕ  r^2 cos^2  ϕ−r^2  cos ϕ sin ϕ+2r^2 sin^2  ϕ−((344)/7)=0  ⇒r^2 =((688)/(7[3−(√2)cos (2ϕ−(π/4))]))  ⇒r=(4/7)(√((301)/(3−(√2)cos (2ϕ−(π/4)))))  ⇒a=r_(max) =(4/7)(√((301)/(3−(√2))))=((4(√(129+43(√2))))/7)  ⇒b=r_(min) =(4/7)(√((301)/(3+(√2))))=((4(√(129−43(√2))))/7)  x=x_m +r cos ϕ=((27)/7)+((4 cos ϕ)/7) (√((301)/(3−(√2)cos (2ϕ−(π/4)))))  y=y_m +r sin ϕ=((19)/7)+((4 sin ϕ)/7) (√((301)/(3−(√2)cos (2ϕ−(π/4)))))  f=2x^2 −3x+4y=f(ϕ)  (df/dϕ)=(4x−3)(dx/dϕ)+4(dy/dϕ)=0  ......  ⇒f_(max) ≈242.6081  ⇒f_(min) ≈−11.4231

$$===\:{An}\:{other}\:{way}\:{to}\:{solve}\:=== \\ $$$${x}^{\mathrm{2}} −\left({y}+\mathrm{5}\right){x}+\mathrm{2}{y}^{\mathrm{2}} −\mathrm{7}{y}−\mathrm{30}=\mathrm{0} \\ $$$$\Delta=\left({y}+\mathrm{5}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{2}{y}^{\mathrm{2}} −\mathrm{7}{y}−\mathrm{30}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{7}{y}^{\mathrm{2}} −\mathrm{38}{y}−\mathrm{145}\leqslant\mathrm{0} \\ $$$${y}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{19}\pm\mathrm{4}\sqrt{\mathrm{86}}}{\mathrm{7}} \\ $$$$\Rightarrow\frac{\mathrm{19}−\mathrm{4}\sqrt{\mathrm{86}}}{\mathrm{7}}\leqslant{y}\leqslant\frac{\mathrm{19}+\mathrm{4}\sqrt{\mathrm{86}}}{\mathrm{7}} \\ $$$$\Rightarrow{y}_{{m}} =\frac{\mathrm{19}}{\mathrm{7}} \\ $$$$\mathrm{2}{y}^{\mathrm{2}} −\left({x}+\mathrm{7}\right){y}+{x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{30}=\mathrm{0} \\ $$$$\Delta=\left({x}+\mathrm{7}\right)^{\mathrm{2}} −\mathrm{8}\left({x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{30}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{7}{x}^{\mathrm{2}} −\mathrm{54}{x}−\mathrm{289}\geqslant\mathrm{0} \\ $$$${x}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{27}\pm\mathrm{8}\sqrt{\mathrm{43}}}{\mathrm{7}} \\ $$$$\Rightarrow\frac{\mathrm{27}−\mathrm{8}\sqrt{\mathrm{43}}}{\mathrm{7}}\leqslant{x}\leqslant\frac{\mathrm{27}+\mathrm{8}\sqrt{\mathrm{43}}}{\mathrm{7}} \\ $$$$\Rightarrow{x}_{{m}} =\frac{\mathrm{27}}{\mathrm{7}} \\ $$$$\Rightarrow{the}\:{points}\:{lie}\:{on}\:{an}\:{ellipse}\:{with} \\ $$$${center}\:{at}\:\left(\frac{\mathrm{27}}{\mathrm{7}},\frac{\mathrm{19}}{\mathrm{7}}\right) \\ $$$${let}\:{x}={u}+\frac{\mathrm{27}}{\mathrm{7}},\:{y}={v}+\frac{\mathrm{19}}{\mathrm{7}} \\ $$$$\left({u}+\frac{\mathrm{27}}{\mathrm{7}}\right)^{\mathrm{2}} −\left({v}+\frac{\mathrm{54}}{\mathrm{7}}\right)\left({u}+\frac{\mathrm{27}}{\mathrm{7}}\right)+\left(\mathrm{2}{v}−\frac{\mathrm{11}}{\mathrm{7}}\right)\left({v}+\frac{\mathrm{19}}{\mathrm{7}}\right)−\mathrm{30}=\mathrm{0} \\ $$$$\Rightarrow{u}^{\mathrm{2}} −{uv}+\mathrm{2}{v}^{\mathrm{2}} −\frac{\mathrm{344}}{\mathrm{7}}=\mathrm{0} \\ $$$${let}\:{u}={r}\:\mathrm{cos}\:\varphi,\:{v}={r}\:\mathrm{sin}\:\varphi \\ $$$${r}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\varphi−{r}^{\mathrm{2}} \:\mathrm{cos}\:\varphi\:\mathrm{sin}\:\varphi+\mathrm{2}{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\varphi−\frac{\mathrm{344}}{\mathrm{7}}=\mathrm{0} \\ $$$$\Rightarrow{r}^{\mathrm{2}} =\frac{\mathrm{688}}{\mathrm{7}\left[\mathrm{3}−\sqrt{\mathrm{2}}\mathrm{cos}\:\left(\mathrm{2}\varphi−\frac{\pi}{\mathrm{4}}\right)\right]} \\ $$$$\Rightarrow{r}=\frac{\mathrm{4}}{\mathrm{7}}\sqrt{\frac{\mathrm{301}}{\mathrm{3}−\sqrt{\mathrm{2}}\mathrm{cos}\:\left(\mathrm{2}\varphi−\frac{\pi}{\mathrm{4}}\right)}} \\ $$$$\Rightarrow{a}={r}_{{max}} =\frac{\mathrm{4}}{\mathrm{7}}\sqrt{\frac{\mathrm{301}}{\mathrm{3}−\sqrt{\mathrm{2}}}}=\frac{\mathrm{4}\sqrt{\mathrm{129}+\mathrm{43}\sqrt{\mathrm{2}}}}{\mathrm{7}} \\ $$$$\Rightarrow{b}={r}_{{min}} =\frac{\mathrm{4}}{\mathrm{7}}\sqrt{\frac{\mathrm{301}}{\mathrm{3}+\sqrt{\mathrm{2}}}}=\frac{\mathrm{4}\sqrt{\mathrm{129}−\mathrm{43}\sqrt{\mathrm{2}}}}{\mathrm{7}} \\ $$$${x}={x}_{{m}} +{r}\:\mathrm{cos}\:\varphi=\frac{\mathrm{27}}{\mathrm{7}}+\frac{\mathrm{4}\:\mathrm{cos}\:\varphi}{\mathrm{7}}\:\sqrt{\frac{\mathrm{301}}{\mathrm{3}−\sqrt{\mathrm{2}}\mathrm{cos}\:\left(\mathrm{2}\varphi−\frac{\pi}{\mathrm{4}}\right)}} \\ $$$${y}={y}_{{m}} +{r}\:\mathrm{sin}\:\varphi=\frac{\mathrm{19}}{\mathrm{7}}+\frac{\mathrm{4}\:\mathrm{sin}\:\varphi}{\mathrm{7}}\:\sqrt{\frac{\mathrm{301}}{\mathrm{3}−\sqrt{\mathrm{2}}\mathrm{cos}\:\left(\mathrm{2}\varphi−\frac{\pi}{\mathrm{4}}\right)}} \\ $$$${f}=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}{y}={f}\left(\varphi\right) \\ $$$$\frac{{df}}{{d}\varphi}=\left(\mathrm{4}{x}−\mathrm{3}\right)\frac{{dx}}{{d}\varphi}+\mathrm{4}\frac{{dy}}{{d}\varphi}=\mathrm{0} \\ $$$$...... \\ $$$$\Rightarrow{f}_{{max}} \approx\mathrm{242}.\mathrm{6081} \\ $$$$\Rightarrow{f}_{{min}} \approx−\mathrm{11}.\mathrm{4231} \\ $$

Commented by mr W last updated on 26/Jan/20

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