Ricerca di massimi e minimi assoluti
Salve a tutti ho un dubbio riguardo il seguente esercizio:
Sia $f(x; y) = y^2x + y-x^2 $ .
Trovare il massimo ed il minimo assoluto (se esistono) di f sul triangolo curvilineo delimitato dalle rette
$x = 0$; $y = 1$ e dalla parabola $y = x^2 $.
Ho disegnato inizialmente il vincolo della mia funzione
[jxg]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[/jxg]
In seguito ho ricavato il gradiente della mia funzione ovvero:
$\grad f(x,y)= (y^2-2x ; 2yx+1)$.
Ho usato quindi il metodo dei moltiplicatori di Lagrange.
Per [size=150]$\gamma$[/size]1: $x=0$;
$\grad \gamma (x,y)= (1,0)$.
$\{(y^2-2x=λ),(2yx+1=0),(x=0):}$ per il quale non esistono soluzioni.
Per Per [size=150]$\gamma$[/size]2: $y=1$;
$\grad \gamma (x,y)= (0,1)$.
$\{(y^2-2x=0),(2yx+1=λ),(y=1):}$ per il quale ottengo il punto $ P (1/2,1) $.
$f(P) = 5/4$ .
Infine per [size=150]$\gamma$[/size]3: $y-x^2=0$;
$\grad \gamma (x,y)= (-2x,1)$.
$\{(y^2-2x=-2xλ),(2yx+1=λ),(y-x^2=0):}$ per il quale ottengo il punto $ O (0,0) $.
Con $f(O)=0$
Dunque come soluzione avrei il solo punto P di massimo assoluto mentre il punto O come minimo assoluto.
Controllando poi su Wolframalpha come unico risultato ottengo un massimo e un minimo locale http://www.wolframalpha.com/input/?i=extrema+y%5E2x%2By%E2%88%92x%5E2+where+x%3D0%2Cy%3D1%2Cy%3Dx%5E2
Cosa ho sbagliato?
Sia $f(x; y) = y^2x + y-x^2 $ .
Trovare il massimo ed il minimo assoluto (se esistono) di f sul triangolo curvilineo delimitato dalle rette
$x = 0$; $y = 1$ e dalla parabola $y = x^2 $.
Ho disegnato inizialmente il vincolo della mia funzione
[jxg]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[/jxg]
In seguito ho ricavato il gradiente della mia funzione ovvero:
$\grad f(x,y)= (y^2-2x ; 2yx+1)$.
Ho usato quindi il metodo dei moltiplicatori di Lagrange.
Per [size=150]$\gamma$[/size]1: $x=0$;
$\grad \gamma (x,y)= (1,0)$.
$\{(y^2-2x=λ),(2yx+1=0),(x=0):}$ per il quale non esistono soluzioni.
Per Per [size=150]$\gamma$[/size]2: $y=1$;
$\grad \gamma (x,y)= (0,1)$.
$\{(y^2-2x=0),(2yx+1=λ),(y=1):}$ per il quale ottengo il punto $ P (1/2,1) $.
$f(P) = 5/4$ .
Infine per [size=150]$\gamma$[/size]3: $y-x^2=0$;
$\grad \gamma (x,y)= (-2x,1)$.
$\{(y^2-2x=-2xλ),(2yx+1=λ),(y-x^2=0):}$ per il quale ottengo il punto $ O (0,0) $.
Con $f(O)=0$
Dunque come soluzione avrei il solo punto P di massimo assoluto mentre il punto O come minimo assoluto.
Controllando poi su Wolframalpha come unico risultato ottengo un massimo e un minimo locale http://www.wolframalpha.com/input/?i=extrema+y%5E2x%2By%E2%88%92x%5E2+where+x%3D0%2Cy%3D1%2Cy%3Dx%5E2
Cosa ho sbagliato?
Risposte
ho avuto lo stesso problema con wolfram alpha
Accedi a tutti gli appunti
Tutor AI: studia meglio e in meno tempo