MHB Solve f(x,y,z): Min & Max Values Subject to Constraint x^2+2y^2+6z^2=81

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The discussion focuses on finding the minimum and maximum values of the function f(x,y,z)=3x+2y+4z under the constraint x^2+2y^2+6z^2=81. The user initially struggles with setting equations equal to each other, resulting in trivial identities. A suggestion is made to express y and z in terms of x and substitute these into the constraint. The user successfully follows this advice, leading to progress in solving the problem. The conversation emphasizes the importance of substitution in optimization problems with constraints.
ajkess1994
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Afternoon, I have been working on this problem for awhile now but have been stuck on a certain point, and once I set everything equal to each other I end up with the same thing for example: x=x, y=y, & z=z

Find the minimum and maximum values of the function f(x,y,z)=3x+2y+4z subject to the constraint x^(2)+2y^(2)+6z^(2)=81.

fmax= ?
fmin= ?

This is what I have,

fx=3, fy=2, fz=4, gx=2x, gy=4y, gz=12z

3=2x(lamda); (lamda)= 3/2x

2=4y(lamda); (lamda)= 2/4y

4=12y(lamda); (lamda)= 4/12z

(3/2x)=(2/4y)=(4/12z)
 
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ajkess1994 said:
Afternoon, I have been working on this problem for awhile now but have been stuck on a certain point, and once I set everything equal to each other I end up with the same thing for example: x=x, y=y, & z=z

Find the minimum and maximum values of the function f(x,y,z)=3x+2y+4z subject to the constraint x^(2)+2y^(2)+6z^(2)=81.

fmax= ?
fmin= ?

This is what I have,

fx=3, fy=2, fz=4, gx=2x, gy=4y, gz=12z

3=2x(lamda); (lamda)= 3/2x

2=4y(lamda); (lamda)= 2/4y

4=12y(lamda); (lamda)= 4/12z

(3/2x)=(2/4y)=(4/12z)

Hi ajkess1994 and welcome to MHB! ;)

Express $y$ in $x$, and also $z$ in $x$, and substitute in $x^2+2y^2+6z^2=81$?
 
ajkess1994 said:
Afternoon, I have been working on this problem for awhile now but have been stuck on a certain point, and once I set everything equal to each other I end up with the same thing for example: x=x, y=y, & z=z

Find the minimum and maximum values of the function f(x,y,z)=3x+2y+4z subject to the constraint x^(2)+2y^(2)+6z^(2)=81.

fmax= ?
fmin= ?

This is what I have,

fx=3, fy=2, fz=4, gx=2x, gy=4y, gz=12z

3=2x(lamda); (lamda)= 3/2x

2=4y(lamda); (lamda)= 2/4y

4=12y(lamda); (lamda)= 4/12z

(3/2x)=(2/4y)=(4/12z)

I would first look at:

$$\frac{3}{2x}=\frac{2}{4y}\implies y=\frac{x}{3}$$

And then:

$$\frac{3}{2x}=\frac{4}{12z}\implies z=\frac{2x}{9}$$

Now, substitute into your constraint, and solve for \(x\)...what do you get?
 
Thank you MarkFL the process carried out and worked.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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