anonymous
  • anonymous
The variables p and q are related by the equation p^3q=9 . Another variable z, is defined by z=16p+3q . Find the value of p and of q that make z a minimum. Just to double check my answer @rational
Mathematics
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SOLVED
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chestercat
  • chestercat
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rational
  • rational
p^3q = 9 solve q from above equation and substitute in the expression you want to minimize
anonymous
  • anonymous
\[p^3q=9-First~eqn\]\[z=16p+3q-Second~eqn\]\[q=\frac{ 9 }{ p^3 }-Third~eqn\]\[z=16p+3(\frac{ 9 }{ p^3 })\]\[z=16p+\frac{ 27 }{ p^3 }\]\[z=16p+27(p^{-3})\]\[\frac{ dz }{ dp }=16-81(p^{-4})\]\[=16-\frac{ 81 }{ p^4 }\]\[when~\frac{ dz }{ dp }=0,\]\[16-\frac{ 81 }{ p^4 }=0\]\[16p^4=81\]\[p^4=\frac{ 81 }{ 16 }\]\[p=\frac{ 3 }{ 2 }\]
anonymous
  • anonymous
i'm getting p=3/2 but the answer say it is p=5/2, i think the book's answer is wrong

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anonymous
  • anonymous
i'm also getting q=8/3 but the answer say it is 3/8
anonymous
  • anonymous
is my answer correct ? @rational
rational
  • rational
that looks good p = -3/2 and q = -8/3 also works right ?
rational
  • rational
\(p^4=\frac{ 81 }{ 16 } \implies p = \pm \dfrac{3}{2}\)
anonymous
  • anonymous
Thank you @rational :)
anonymous
  • anonymous
so the answer is \[p=\pm \frac{ 3 }{ 2 }and~q=\pm \frac{ 8 }{ 3 }\]
rational
  • rational
Yep!

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