0
$\begingroup$

Find three positive integers x, y, z that satisfy the given conditions. The product is 27, and the sum is a minimum.

I'm lost on how I would solve my system of equations that I have set up.

$1=\lambda \cdot yz$

$1=\lambda \cdot xz$

$1=\lambda \cdot xy$

$xyz=27$

Are my system of equations correct?

$\endgroup$
1
$\begingroup$

Yes, the system is$$\left\{\begin{array}{l}1=\lambda yz\\1=\lambda xz\\1=\lambda xy\\xyz=27.\end{array}\right.$$It follows from any of the first $3$ equations that none of the numbers $x$, $y$, and $z$ can be $0$. So$$1=\frac11=\frac{\lambda yz}{\lambda xz}=\frac yx;$$that is, $y=x$. By the same argument, $z=x=y$. Can you take it from here?

$\endgroup$
  • $\begingroup$ yes, thank you so much! Really appreciate your response. $\endgroup$ – Ian Sid Oct 24 '20 at 19:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.