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- #1

#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

Given:

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

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

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

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

- Thread starter Albert
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- Thread starter
- #1

- Jan 25, 2013

- 1,225

Given:

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

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

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

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

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- #2

My solution:

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

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:

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

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- #3

- Jan 25, 2013

- 1,225

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

I will give you an example :

(x,y,z)=(79, 2, 2)

then xy+z=160

x+yz=83<160

Last edited:

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- #4

I do find that the Second Partials Test reveals that the critical point is a saddle point.sorry the answer is not correct

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

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

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- #5

- Feb 7, 2012

- 2,786

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

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- #6

- Jan 25, 2013

- 1,225

the answer is correct ,try to solve it systematically pleaseBest I can do so far (by trial and error) is $(x,y,z) = (26,6,4)$, giving $x+yz = 50$.

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- #7

- Jan 25, 2013

- 1,225

$(xy+z=160)$

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