Word problem using lagrange multiplier

[jonroberts74]

Homework Statement

The Baraboo, Wisconsin plant of International Widget Co. uses aluminum, iron and magnesium to produce high-quality widgets. The quantity of widgets which may be produced using x tonnes of aluminum, y tonnes of iron and z tonnes of magnesium is Q(x,y,z) = xyz. the cost of raw materials is aluminum, $6 dollars per tonne; iron, $4 per tonne; magnesium, $8 per tonne. How many of each of aluminum, iron, and magnesium should be used to manufacture 1000 widgets at the lowest possible cost

Homework Equations

Q(x,y,z) = xyz is the objective function

is the constraint 6x+4y+8z=1000 ?

The Attempt at a Solution

 

[vela]

No, it isn't. How did you come up with that? What is 6x+4y+8z supposed to represent? What does the 1000 represent? Does it make sense to set those two expressions equal to each other?

 

[jonroberts74]

I have issues with reading english.

the 1000 represents the number of widgets

6x,4y,8z would be the cost of each used material.

now I see the issue because 1000 is not a cost, it is a quantity

1000=xyz

so I need to find my objective function

would it not be the cost of each added up?
6x+4y+8z=C(x,y,z) [doesn't make much sense for solving it either though when I try to actually try putting it together]

 

[vela]

Yup, you're trying to minimize the cost subject to the constraint of a fixed quantity.

 

[Ray Vickson]

I have issues with reading english.

the 1000 represents the number of widgets

6x,4y,8z would be the cost of each used material.

now I see the issue because 1000 is not a cost, it is a quantity

1000=xyz

so I need to find my objective function

would it not be the cost of each added up?
6x+4y+8z=C(x,y,z) [doesn't make much sense for solving it either though when I try to actually try putting it together]

Yes, your problem is to minimize 6x+4y+8z, subject to the constraint that xyz = 1000. Implicitly, you also need x,y,z > 0.

The problem has a perfectly respectable solution that can be found via a Lagrange multiplier method, or by using the constraint to solve for z, say, in terms of x and y, then using that in the objective to get an unconstrained problem (although still with implicit constraints x,y > 0).

 

[jonroberts74]

thank you for the help, word problems are hard with my difficulty reading



F(x,y,z,\lambda) = 6x+4y+8z - \lambda(xyz-1000)

\left\{\begin{array}{cc}6-yz\lambda=0 \\ 4-xz\lambda=0 \\8-xy\lambda=0 \\ -xyz+1000 =0 \end{array}\right.

\lambda = \frac{6}{yz} = \frac{4}{xz} = \frac{8}{xy}

whats the best step to take from here

 

[jonroberts74]

thank you for the help, word problems are hard with my difficulty reading



F(x,y,z,\lambda) = 6x+4y+8z - \lambda(xyz-1000)

\left\{\begin{array}{cc}6-yz\lambda=0 \\ 4-xz\lambda=0 \\8-xy\lambda=0 \\ -xyz+1000 =0 \end{array}\right.

\lambda = \frac{6}{yz} = \frac{4}{xz} = \frac{8}{xy}

whats the best step to take from here

If I were able to isolate each equality for lambda to one variable then I could solve the restraint for lambda to get the x,y,z. and it seems if I try solving for any variable it will have two variable in the equality

 

[vela]

What have you tried? It seems pretty straightforward how to solve this system.

 

[jonroberts74]

I tried solving for lambda first but that doesn't work.

Then tried solving for z, but I didn't have luck. I'll try again this afternoon.

 

[vela]

Show your work if you want help. Simply saying what you tried and then saying it didn't work is pretty much useless to us.

 

[jonroberts74]

I can't show it when I am replying in between classes. That why I said I will do it again this afternoon.

 

[jonroberts74]

F(x,y,z,\lambda) = 6x+4y+8z - \lambda(xyz-1000)

\left\{\begin{array}{cc}6-yz\lambda=0 \\ 4-xz\lambda=0 \\8-xy\lambda=0 \\ -xyz+1000 =0 \end{array}\right.

\lambda = \frac{6}{yz} = \frac{4}{xz} = \frac{8}{xy}

if I take

6-yz\lambda=0; 4-xz\lambda=0\lambda = \frac{6}{yz} = \frac{4}{zx}

solving for x x=\frac{2y}{3} now using

8-xy\lambda=0;4-xz\lambda=0 \Rightarrow \lambda = \frac{8}{xy}=\frac{4}{xz}

solving for z z=\frac{1y}{2}

now \frac{2y}{3}(y)\frac{1y}{2}=1000\Rightarrow \frac{1y}{3}=1000 \Rightarrow y =3000

x=2000;z=1500
this seems incorrect

 

[vela]

$\frac 23 y \times y \times \frac 12 y = \frac 13 y^3$, not $\frac 13 y$.

 

[jonroberts74]

$\frac 23 y \times y \times \frac 12 y = \frac 13 y^3$, not $\frac 13 y$.

that's how tried I am. sorry

so \frac{y^3}{3}=1000 \Rightarrow y=\sqrt[3]{1000/3}

 

[vela]

Not quite. Try again.

 

[jonroberts74]

I see what's wrong

y= 10\sqrt[3]{3}

 

[jonroberts74]

y= 10\sqrt[3]{3}; x = \frac{20\sqrt[3]{3}}{3};z=5\sqrt[3]{3}

 

[vela]

Looks good!