2

I have:

$$f(x,y,z) = x + y + z$$

and given the constraints: $\displaystyle \frac a x + \frac b y + \frac c z = 1$, and $x, y, z > 0$, where $a,b,c$ are fixed positive constants.

I need to find the minimum value of $f(x, y, z)$.

I have taken partial derivatives and tried using Lagrange Multipliers but am stuck:

$$1 = - \dfrac {\lambda a}{x^2}; \\ 1 = - \dfrac {\lambda b}{y^2}; \\ 1 = - \dfrac {\lambda c}{z^2};$$

I'm not really sure how to use the fact that $x,y,z >0$ and $a,b,c$ are fixed positive constants.

mookid
  • 28,236
lllll
  • 409

2 Answers2

0

Follow your nose for a while: Assuming an extreme exists,

$$x = \sqrt{-\lambda a}, y = \sqrt{-\lambda b}, z = \sqrt{-\lambda c}$$

Substituting back into the constraint equation

$$\frac{1}{\sqrt{-\lambda}} \left( \sqrt{a} + \sqrt{b} + \sqrt{c} \right) = 1$$

You can now solve for $\lambda$ and hence $x, y, z$. Then check to see if this corresponds to a min or max for $f$.

Simon S
  • 26,524
0

Tip: the only function of $\lambda $ is to disappear. So, the conclusion of your Lagrange equations is that $$ \frac {x^2}a =\frac {y^2}b =\frac {z^2}c $$

With the constrain, you should be able to take it back to a one variable optimization problem.

mookid
  • 28,236
  • Sorry, could you elaborate on your answer. How would I turn it into a one variable optimization problem? – lllll Nov 05 '14 at 00:33
  • @lllll using this equation and the constrain, you have two equations. Using these, solve to get $y,z$ as functions of $x$ and put everything back in the thing to optimize. – mookid Nov 05 '14 at 00:35
  • I solved for y and z in terms of x and plugged it into the constraint and got a/x + sqrt(ab)/x + sqrt(ac)/x = 1 and i solved for x. do I do the same for y and z and that will be my minimum point in terms of a,b,c? – lllll Nov 05 '14 at 01:26
  • exactly. if your computations are right this should yield the right result. – mookid Nov 05 '14 at 01:39