# Word problem using lagrange multiplier

1. Jul 13, 2014

### jonroberts74

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

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

3. The attempt at a solution

Last edited: Jul 13, 2014
2. Jul 13, 2014

### vela

Staff Emeritus
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?

3. Jul 13, 2014

### 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]

Last edited: Jul 13, 2014
4. Jul 13, 2014

### vela

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

5. Jul 13, 2014

### Ray Vickson

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).

6. Jul 13, 2014

### 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

Last edited: Jul 13, 2014
7. Jul 16, 2014

### jonroberts74

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

8. Jul 16, 2014

### vela

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

9. Jul 16, 2014

### 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.

10. Jul 16, 2014

### vela

Staff Emeritus
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.

11. Jul 16, 2014

### jonroberts74

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

12. Jul 16, 2014

### 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

13. Jul 16, 2014

### vela

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

14. Jul 16, 2014

### jonroberts74

that's how tried I am. sorry

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

15. Jul 16, 2014

### vela

Staff Emeritus
Not quite. Try again.

16. Jul 16, 2014

### jonroberts74

I see whats wrong

$$y= 10\sqrt[3]{3}$$

Last edited: Jul 16, 2014
17. Jul 16, 2014

### jonroberts74

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

18. Jul 16, 2014

### vela

Staff Emeritus
Looks good!