Question 1050599
{{{system(xy=3, p=2x+2y)}}}


{{{y=3/x}}}
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{{{p=2x+2(3/x)}}}
{{{p=2x+6/x}}}
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{{{dp/dx=2+6(-1)x^(-2)}}}
{{{dp/dx=2-6/x^2}}}
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Any minimum of maximum for p occurs for {{{dp/dx=0}}}.
{{{2-6/x^2=0}}}
{{{2=6/x^2}}}
{{{x^2/6=1/2}}}
{{{x^2=6/2}}}
{{{x^2=3}}}
{{{x=0+- sqrt(3)}}}
You must use a positive real value, so {{{highlight(x=sqrt(3))}}}.


y, the other dimension:
{{{y=x/3}}}
{{{highlight(y=sqrt(3)/3)}}}



Just a technological graphical check if this really is a minimum:
{{{p=2x+2(3/x)}}}
{{{p=2x+6/x}}}
{{{graph(300,300,-1,15,-1,15,2x+6/x)}}}