Find the constrained maxima and minima of ##f(x,y,z)=x+y^2+2z##

[chwala]

Homework Statement

See attached- My interest is on number 4.

Relevant Equations

$\nabla f=0$ and Lagrange multiplier.

The interest is on number $4$,
In my working,

$f(x,y,z) = x+y^2+2z$ and $g(x,y,z) = 4x^2+9y^2-36z^2 = 36$

$f_x = 1, f_y=2y$ and $f_z = 2$ and also $g_x = 8λx, g_y = 18λy$ and $g_z = -72λz$

using $\nabla f (x,y,z) = λ\nabla f (x,y,z)$

i shall have,

$1 = 8λx$

$2y = 18λy$

$2 = -72λz$

then,

$λ =\dfrac{1}{8x} = \dfrac{1}{9}= \dfrac{-1}{36z}$

$\dfrac{1}{9}= \dfrac{-1}{36z}$

$z= \dfrac{-1}{4}$ and we also have, $\dfrac{1}{8x} = \dfrac{1}{9} ⇒ x = \dfrac{9}{8}$

We have $4x^2+9y^2-36z^2=36$

To find $y$,

⇒$4⋅\left(\dfrac{9}{8}\right)^2 + 9y^2 - 36 ⋅\left(\dfrac{-1}{4}\right)^2=36$

$576y^2 = 2124$

$y^2 = \dfrac{59}{\sqrt{16}}$

$y = \dfrac{\sqrt{59}}{4}$

$(x,y,z) = \left(\dfrac{9}{8}, \dfrac{\sqrt{59}}{4},\dfrac{-1}{4} \right)$

also when $y=0$ from the equation, $2y = 18λy$

we shall have with similar steps $(x,y,z) = \left(\dfrac{-9}{\sqrt{5}}, 0, \dfrac{2}{\sqrt{5}}\right)$

any input or alternative welcome...

Bingo!

 

[pasmith]

Homework Statement: See attached- My interest is on number 4.
Relevant Equations: $\nabla f=0$ and Lagrange multiplier.

View attachment 336621

The interest is on number $4$,
In my working,

$f(x,y,z) = x+y^2+2z$ and $g(x,y,z) = 4x^2+9y^2-36z^2 = 36$

This should be $g(x,y,z) = 4x^2 + 9y^2 - 36z^2 - 36.$ Note that $y$ appears in $f$ and $g$ only as $y^2$, so if $(x,y,z)$ is a constrained extremum then so is $(x,-y,z)$.

$f_x = 1, f_y=2y$ and $f_z = 2$ and also $g_x = 8λx, g_y = 18λy$ and $g_z = -72λz$

using $\nabla f (x,y,z) = λ\nabla f (x,y,z)$

You mean $\nabla f = \lambda \nabla g$.

i shall have,

$1 = 8λx$

$2y = 18λy$

$2 = -72λz$

then,

$λ =\dfrac{1}{8x} = \dfrac{1}{9}= \dfrac{-1}{36z}$

So we are assuming here that $9\lambda = 1$.

$\dfrac{1}{9}= \dfrac{-1}{36z}$

$z= \dfrac{-1}{4}$ and we also have, $\dfrac{1}{8x} = \dfrac{1}{9} ⇒ x = \dfrac{9}{8}$

We have $4x^2+9y^2-36z^2=36$

To find $y$,

⇒$4⋅\left(\dfrac{9}{8}\right)^2 + 9y^2 - 36 ⋅\left(\dfrac{-1}{4}\right)^2=36$

$576y^2 = 2124$

$y^2 = \dfrac{59}{\sqrt{16}}$

$y = \dfrac{\sqrt{59}}{4}$

$(x,y,z) = \left(\dfrac{9}{8}, \dfrac{\sqrt{59}}{4},\dfrac{-1}{4} \right)$

also when $y=0$ from the equation, $2y = 18λy$

we shall have with similar steps $(x,y,z) = \left(\dfrac{-9}{\sqrt{5}}, 0, \dfrac{2}{\sqrt{5}}\right)$

For $y = 0$ you need $$4x^2 - 36z^2 = 4\frac{1}{(8\lambda)^2} - 36\frac{1}{(36\lambda)^2} = 36.$$ That gives two solutions for $\lambda$; you have shown only one result.

 

[chwala]

This should be $g(x,y,z) = 4x^2 + 9y^2 - 36z^2 - 36.$ Note that $y$ appears in $f$ and $g$ only as $y^2$, so if $(x,y,z)$ is a constrained extremum then so is $(x,-y,z)$.
You mean $\nabla f = \lambda \nabla g$.
So we are assuming here that $9\lambda = 1$.
For $y = 0$ you need $$4x^2 - 36z^2 = 4\frac{1}{(8\lambda)^2} - 36\frac{1}{(36\lambda)^2} = 36.$$ That gives two solutions for $\lambda$; you have shown only one result.

Let me check on the second result.