Compute min(x+yz)

[Albert]

Given:

$x,y,z\in\mathbb{N}\text{ and }xy+z=160$

$\text{Compute }\min(x+yz)$

 

[MarkFL]

Re: compute min(x+yz)

My solution:

Spoiler

We have the objective function:

\(\displaystyle f(x,y,z)=x+yz\)

subject to the constraint:

\(\displaystyle g(x,y,z)=xy+z-160=0\)

Using Lagrange multipliers, we find:

\(\displaystyle 1=\lambda y\)

\(\displaystyle z=\lambda x\)

\(\displaystyle y=\lambda\)

This, along with the constraint, implies:

\(\displaystyle y=1,\,x=z\)

and so the constraint gives us:

\(\displaystyle x=z=80\)

The testing of $(x,y,z)=(40,2,80)$ reveals that the critical point is a minimum. Hence:

\(\displaystyle f_{\min}=f(80,1,80)=160\)

 

[Albert]

sorry the answer is not correct
I will give you an example :
(x,y,z)=(79, 2, 2)
then xy+z=160
x+yz=83<160

 

[MarkFL]

sorry the answer is not correct

I do find that the Second Partials Test reveals that the critical point is a saddle point.

If I had checked the boundaries, for example $(x,y,z)=(3,53,1)$, I would find:

\(\displaystyle f(x,y,z)=56\).

 

[Opalg]

Best I can do so far (by trial and error) is $(x,y,z) = (26,6,4)$, giving $x+yz = 50$.

 

[Albert]

Best I can do so far (by trial and error) is $(x,y,z) = (26,6,4)$, giving $x+yz = 50$.

the answer is correct ,try to solve it systematically please

 

[Albert]

$x,y,z \in N $
$(xy+z=160)$
let :$y=\dfrac {160-z}{x}$ , yielding $ x + yz = x + \dfrac{z(160−z)}{x}$ .
Thus $x + yz ≥ 2\sqrt{xyz}=2\sqrt{ z(160 − z)}$
Considered as a function of z
now we begin with small values of z, and (160 − z)
such that x is close to $\sqrt{z(160 − z).}$
(Equivalently, such that :$\dfrac{x}{y}$ is close to z. or x is close to yz)
this will happen for (x,y,z)=(26,6,4)
(here 26 is close to $yz=6\times 4$ )
and we have the minimal value of x + yz is 50