At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

what are the equations?

what is the function more importantly.

f(x,y,z)=x+z^2
g(x,y,z)=x^2+y^2
h(x,y,z)=2y+z

Maximize f(x,y,z) subject to the restriction g and h?

actually i know how to solve

but are these g and h related or both different

yeah maximize with restriction g and h i think

I would let G(x,y,z) = ( x^2 + y^2 , 2y + z)
where this our constrained set.

so now partial derivative of f(x)=lamba(g(x))

yes,

\[\frac{ part( f}{par( x,y,z)} = (\lambda_1, \lambda_2, \lambda_3)\frac{par G}{par (x,y,z)}\]

if its two dimensional we need to take only 1 lambda then why 3 for 3D?

I don't know how you need 1 eigenvalue for 2. You need 2.

lambda was probably written as a vector.

and it had two components.

but here its 3D and i dont know how to solve this

Same business here.

your part(g)/part(x,yz) is going to be a matrix 3x3.

find its eigenvalues using characteristic polynomial

or however you find eigenvalues.

but here f(x,y,z)=lambda g(x,y,z)or f(x,y,z)=lambdah(x,y,z)
how i beed to proceed

W(x,y,z) = ( x^2 + y^2 , 2y + z)

| 2x 2y 0 |
| 0 2y 1 |
| |

ok i will try to solve and wil you be available online??

what do you get as partial f(x,y,z)

<1,0,2z>

<1,0,2z> = | 2x 2y 0 |
| 0 2y 1 |

here 3 equations and 5 unknowns

Don't worry we are just trying to find expressions for x,y,z not explicit answers.
what did you get?

i got lam1=1/2x
2y(lam1+lam2)=0
lam2=2z

try to get expressions for x, y,z and remember you can still use you're functions g, h

BTW, are you sure the functions weren't equal to anything?

no

g(x,y,z)=x^2+y^2
h(x,y,z)=2y+z
These two can describe tons of functions.

if i solve i got 2 points (1,0,0) and (-1,0,0)
are these correct??

or are we doing a general case? set them equal to x^2 + y^2 = a
2y+z = b.
How did you do that..

i solved the above equations with g(x) ,h(x)

x^2 + y^2 = 1 and 2y+z = 0 with the equations with lambdas

oh so they were equal to something.

COOL! i was getting scared.

I hope so, don't make me do the algebra =P

x^2 + y^2 = 1 and 2y+z = 0 was this given?

yes