Find the maximum of $f(x,y,z)=x^2y^2z$ in $D$.
$D=\{(x,y,z)\in\mathbb{R}^3| x+y+z\le 5, x,y,z \ge 0 \}$.
My work:
Trying to find inner maximum points: $\nabla f = (2xy^2z, 2yx^2z, x^2y^2)=0$.
I'm not sure how to solve this to find points..
Next step is to check the borders:
$x+y+z=5 \Longrightarrow z=5-x-y$.
So we get: $g(x,y) = x^2y^2(5-x-y)=5x^2y^2 - x^3y^2-x^2y^3=0$.
$g_x=10xy^2+3x^2y^2-2xy^3=y^2(10x+3x^2-2xy)=0$ so I got a point $y=0$.. But I don't know how to solve the other part of the equation.
I'm not really sure if it's my weak algebra that's making me fail to do this question or it's bad decisions of me trying to solve it this way, I would really appreciate any help and hints!
$f(x,y,z)=x^2y^2z$
$D=\{(x,y,z)\in\mathbb{R}^3| x+y+z\le 5, x,y,z \ge 0 \}$
Hint: As $x, y, z \ge 0$, $f(x,y,z)=x^2y^2z$ will be maximum when $x+y+z = 5$ and $x, y, z \ne 0$.
So applying Lagrange Multiplier method,
$2xy^2z = \lambda$
$2x^2yz = \lambda$
$x^2y^2 = \lambda$
If $x, y, z \ne 0$, equating first and second, we get $x = y$ and from second and third, $z = \frac{x}{2} = \frac{y}{2}$. Plug them in $x+y+z = 5$ to find value of $x, y, z$.
Can you take it from here?