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

chwala
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Homework Statement
See attached- My interest is on number 4.
Relevant Equations
##\nabla f=0## and Lagrange multiplier.
1701686538682.png


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!
 
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chwala said:
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.
 
pasmith said:
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.
 
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