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 [itex]g(x,y,z) = 4x^2 + 9y^2 - 36z^2 - 36.[/itex] Note that [itex]y[/itex] appears in [itex]f[/itex] and [itex]g[/itex] only as [itex]y^2[/itex], so if [itex](x,y,z)[/itex] is a constrained extremum then so is [itex](x,-y,z)[/itex].

##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 [itex]\nabla f = \lambda \nabla g[/itex].

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 [itex]9\lambda = 1[/itex].

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

 

[chwala]

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

Let me check on the second result.