# Compute min(x+yz)

#### Albert

##### Well-known member
Given:

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

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

#### MarkFL

Staff member
Re: compute min(x+yz)

My solution:

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

##### Well-known member
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

Last edited:

#### MarkFL

Staff member
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

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

#### Albert

##### Well-known member
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

##### Well-known member
$x,y,z \in N$
$(xy+z=160)$
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