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Question Number 203526 by Mastermind last updated on 21/Jan/24
Determine the maximum and minimum of the function: f(x,y)=x^4 +4x^2 y^2 −2x^2 +2y^2 −1 Thank you
$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{and}\:\mathrm{minimum}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{function}: \\ $$$$\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{x}^{\mathrm{4}} +\mathrm{4x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} −\mathrm{2x}^{\mathrm{2}} +\mathrm{2y}^{\mathrm{2}} −\mathrm{1} \\ $$$$ \\ $$$$\mathrm{Thank}\:\mathrm{you} \\ $$
Answered by aleks041103 last updated on 21/Jan/24
let us see for which c∈R, ∃(x,y):f(x,y)=c x^2 =a, y^2 =b ⇒f=a^2 +4ab−2a+2b−1=c (a^2 −2a−1)+2b(2a+1)=c ⇒b=((−a^2 +2a+c+1)/(2a+1))=y^2 ≥0 also a=x^2 ≥0 −a^2 +2a+c+1=0⇒a_(1,2) =((−2±(√(8+4c)))/(−2))=1±(√(2+c)) 2a+1=0 ⇒ a_3 =−1/2 ⇒a≥0, 2a+1>0 1st case: c<−2 ⇒ −a^2 +2a+1+c<0 ⇒b<0→contradiction 2nd case: c=−2 ⇒ −a^2 +2a+1+c≤0 where for a=1 and b=0 and for a>0, b<0 → contradict 3rd case: −1≥c>−2 ⇒b≥0 for 1−(√(2+c))≤a≤1+(√(2+c)) 4th case: c>−1 ⇒b≥0 for 0≤a≤1+(√(2+c)) ⇒ for all c≥−2, ∃(x,y):f(x,y)=c ⇒min of f(x,y) at { ((x=±1)),((y=0)) :} with min(f)=−2 ∄max of f(x,y)
$${let}\:{us}\:{see}\:{for}\:{which}\:{c}\in\mathbb{R},\:\exists\left({x},{y}\right):{f}\left({x},{y}\right)={c} \\ $$$${x}^{\mathrm{2}} ={a},\:{y}^{\mathrm{2}} ={b} \\ $$$$\Rightarrow{f}={a}^{\mathrm{2}} +\mathrm{4}{ab}−\mathrm{2}{a}+\mathrm{2}{b}−\mathrm{1}={c} \\ $$$$\left({a}^{\mathrm{2}} −\mathrm{2}{a}−\mathrm{1}\right)+\mathrm{2}{b}\left(\mathrm{2}{a}+\mathrm{1}\right)={c} \\ $$$$\Rightarrow{b}=\frac{−{a}^{\mathrm{2}} +\mathrm{2}{a}+{c}+\mathrm{1}}{\mathrm{2}{a}+\mathrm{1}}={y}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${also}\:{a}={x}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$−{a}^{\mathrm{2}} +\mathrm{2}{a}+{c}+\mathrm{1}=\mathrm{0}\Rightarrow{a}_{\mathrm{1},\mathrm{2}} =\frac{−\mathrm{2}\pm\sqrt{\mathrm{8}+\mathrm{4}{c}}}{−\mathrm{2}}=\mathrm{1}\pm\sqrt{\mathrm{2}+{c}} \\ $$$$\mathrm{2}{a}+\mathrm{1}=\mathrm{0}\:\Rightarrow\:{a}_{\mathrm{3}} =−\mathrm{1}/\mathrm{2} \\ $$$$\Rightarrow{a}\geqslant\mathrm{0},\:\mathrm{2}{a}+\mathrm{1}>\mathrm{0} \\ $$$$ \\ $$$$\mathrm{1}{st}\:{case}:\:{c}<−\mathrm{2}\:\Rightarrow\:−{a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{1}+{c}<\mathrm{0} \\ $$$$\Rightarrow{b}<\mathrm{0}\rightarrow{contradiction} \\ $$$$ \\ $$$$\mathrm{2}{nd}\:{case}:\:{c}=−\mathrm{2}\:\Rightarrow\:−{a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{1}+{c}\leqslant\mathrm{0} \\ $$$${where}\:{for}\:{a}=\mathrm{1}\:{and}\:{b}=\mathrm{0} \\ $$$${and}\:{for}\:{a}>\mathrm{0},\:{b}<\mathrm{0}\:\rightarrow\:{contradict} \\ $$$$ \\ $$$$\mathrm{3}{rd}\:{case}:\:−\mathrm{1}\geqslant{c}>−\mathrm{2} \\ $$$$\Rightarrow{b}\geqslant\mathrm{0}\:{for}\:\mathrm{1}−\sqrt{\mathrm{2}+{c}}\leqslant{a}\leqslant\mathrm{1}+\sqrt{\mathrm{2}+{c}} \\ $$$$ \\ $$$$\mathrm{4}{th}\:{case}:\:{c}>−\mathrm{1} \\ $$$$\Rightarrow{b}\geqslant\mathrm{0}\:{for}\:\mathrm{0}\leqslant{a}\leqslant\mathrm{1}+\sqrt{\mathrm{2}+{c}} \\ $$$$ \\ $$$$\Rightarrow\:{for}\:{all}\:{c}\geqslant−\mathrm{2},\:\exists\left({x},{y}\right):{f}\left({x},{y}\right)={c} \\ $$$$ \\ $$$$\Rightarrow{min}\:{of}\:{f}\left({x},{y}\right)\:{at}\:\begin{cases}{{x}=\pm\mathrm{1}}\\{{y}=\mathrm{0}}\end{cases}\:{with}\:{min}\left({f}\right)=−\mathrm{2} \\ $$$$\nexists{max}\:{of}\:{f}\left({x},{y}\right) \\ $$
Answered by AST last updated on 21/Jan/24
f(x,y): (x^2 −1)^2 +(2xy)^2 +(y(√2))^2 −2≥−2 (Equality:x=+_− 1,y=0) No maximum,f(x,y)→∞ as ∣x∣,∣y∣→∞
$${f}\left({x},{y}\right):\:\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}{xy}\right)^{\mathrm{2}} +\left({y}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{2}\geqslant−\mathrm{2}\: \\ $$$$\left({Equality}:{x}=\underset{−} {+}\mathrm{1},{y}=\mathrm{0}\right) \\ $$$${No}\:{maximum},{f}\left({x},{y}\right)\rightarrow\infty\:{as}\:\mid{x}\mid,\mid{y}\mid\rightarrow\infty \\ $$

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