Mixing
Maximum of a two variable function within a defined domain
The function is this: $xye^{frac{(x+y^2)}{4}}$ in the domain $x+ygeq1, ygeq0$
I have done the partial derivatives to get the stationary point $(0,0)$ but it is not a maximum, it is a saddle point. I graphed the domain and it is the whole plane from $(1,0)$.
Attempted solution
The questions says that there is a maximum somewhere in the domain; I must find it.
Since $ygeq0$ I attempted to evaluate the function in $f(x,0)=0$ which gives $x=0$ and therefore gives $f=0$, plus correct me if I am wrong but $x=0$ is out of the domain. I did the same for $f(1,y)=0$ and gives $y=0$. But neither $(1,0)$ nor $(0,1)$ seem as a maximum point to me. Then I, ignorantly and with my poor mathematical skills, improvised and evaluated when $f(x,1-x)$ and $f(1-y,y)$ because this is implicitly what the domain restriction says $(x+ygeq1)$ and somehow I got a point in $(1,1)$ if that is even possible. I tried to prove it with the determinant of the hessian matrix but calculating the second partial derivatives and then the determinant is way too complicated and long for an exam question.
Is $(1,1)$ the point which maximizes the function in this domain?
calculus multivariable-calculus optimization maxima-minima hessian-matrix
edited Nov 21 '18 at 6:13
Yadati Kiran
1,693619
asked Nov 21 '18 at 3:38
Escribas
11
add a comment |
The function is this: $xye^{frac{(x+y^2)}{4}}$ in the domain $x+ygeq1, ygeq0$
I have done the partial derivatives to get the stationary point $(0,0)$ but it is not a maximum, it is a saddle point. I graphed the domain and it is the whole plane from $(1,0)$.
Attempted solution
The questions says that there is a maximum somewhere in the domain; I must find it.
Since $ygeq0$ I attempted to evaluate the function in $f(x,0)=0$ which gives $x=0$ and therefore gives $f=0$, plus correct me if I am wrong but $x=0$ is out of the domain. I did the same for $f(1,y)=0$ and gives $y=0$. But neither $(1,0)$ nor $(0,1)$ seem as a maximum point to me. Then I, ignorantly and with my poor mathematical skills, improvised and evaluated when $f(x,1-x)$ and $f(1-y,y)$ because this is implicitly what the domain restriction says $(x+ygeq1)$ and somehow I got a point in $(1,1)$ if that is even possible. I tried to prove it with the determinant of the hessian matrix but calculating the second partial derivatives and then the determinant is way too complicated and long for an exam question.
Is $(1,1)$ the point which maximizes the function in this domain?
calculus multivariable-calculus optimization maxima-minima hessian-matrix
edited Nov 21 '18 at 6:13
Yadati Kiran
1,693619
asked Nov 21 '18 at 3:38
Escribas
11
add a comment |
The function is this: $xye^{frac{(x+y^2)}{4}}$ in the domain $x+ygeq1, ygeq0$
I have done the partial derivatives to get the stationary point $(0,0)$ but it is not a maximum, it is a saddle point. I graphed the domain and it is the whole plane from $(1,0)$.
Attempted solution
The questions says that there is a maximum somewhere in the domain; I must find it.
Since $ygeq0$ I attempted to evaluate the function in $f(x,0)=0$ which gives $x=0$ and therefore gives $f=0$, plus correct me if I am wrong but $x=0$ is out of the domain. I did the same for $f(1,y)=0$ and gives $y=0$. But neither $(1,0)$ nor $(0,1)$ seem as a maximum point to me. Then I, ignorantly and with my poor mathematical skills, improvised and evaluated when $f(x,1-x)$ and $f(1-y,y)$ because this is implicitly what the domain restriction says $(x+ygeq1)$ and somehow I got a point in $(1,1)$ if that is even possible. I tried to prove it with the determinant of the hessian matrix but calculating the second partial derivatives and then the determinant is way too complicated and long for an exam question.
Is $(1,1)$ the point which maximizes the function in this domain?
calculus multivariable-calculus optimization maxima-minima hessian-matrix
edited Nov 21 '18 at 6:13
Yadati Kiran
1,693619
asked Nov 21 '18 at 3:38
Escribas
11
The function is this: $xye^{frac{(x+y^2)}{4}}$ in the domain $x+ygeq1, ygeq0$
I have done the partial derivatives to get the stationary point $(0,0)$ but it is not a maximum, it is a saddle point. I graphed the domain and it is the whole plane from $(1,0)$.
Attempted solution
The questions says that there is a maximum somewhere in the domain; I must find it.
Since $ygeq0$ I attempted to evaluate the function in $f(x,0)=0$ which gives $x=0$ and therefore gives $f=0$, plus correct me if I am wrong but $x=0$ is out of the domain. I did the same for $f(1,y)=0$ and gives $y=0$. But neither $(1,0)$ nor $(0,1)$ seem as a maximum point to me. Then I, ignorantly and with my poor mathematical skills, improvised and evaluated when $f(x,1-x)$ and $f(1-y,y)$ because this is implicitly what the domain restriction says $(x+ygeq1)$ and somehow I got a point in $(1,1)$ if that is even possible. I tried to prove it with the determinant of the hessian matrix but calculating the second partial derivatives and then the determinant is way too complicated and long for an exam question.
Is $(1,1)$ the point which maximizes the function in this domain?
calculus multivariable-calculus optimization maxima-minima hessian-matrix
calculus multivariable-calculus optimization maxima-minima hessian-matrix
edited Nov 21 '18 at 6:13
Yadati Kiran
1,693619
asked Nov 21 '18 at 3:38
Escribas
11
edited Nov 21 '18 at 6:13
Yadati Kiran
1,693619
asked Nov 21 '18 at 3:38
Escribas
11
edited Nov 21 '18 at 6:13
Yadati Kiran
1,693619
edited Nov 21 '18 at 6:13
Yadati Kiran
1,693619
edited Nov 21 '18 at 6:13
Yadati Kiran
1,693619
1,693619
asked Nov 21 '18 at 3:38
Escribas
11
asked Nov 21 '18 at 3:38
Escribas
11
asked Nov 21 '18 at 3:38
Escribas
11
11
add a comment |
add a comment |
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