Question

Hi,
Can someone help me with this problem?
Thanks in advance!

Answer 1

`\bf QP^{(1)/(2)}=38\implies \stackrel{\textit{product rule}}{\cfrac{dQ}{dP}P^{(1)/(2)}+Q\cdot \cfrac{1}{2}P^{-(1)/(2)}}=0\implies \cfrac{dQ}{dP}P^{(1)/(2)}=-\cfrac{Q}{2P^{(1)/(2)}} \\\\\\ \cfrac{dQ}{dP}=-\cfrac{Q}{2P^{(1)/(2)}P^{(1)/(2)}}\implies \cfrac{dQ}{dP}=-\cfrac{Q}{2P}\impliedby \textit{now, let's zero it out} \\\\\\ \stackrel{\textit{horizontal tangent line}}{0=-\cfrac{Q}{2P}}\implies 0=-Q\implies 0=Q`

now, how to check it? well, graphing the equation, to check where it may have an extrema, namely a horizontal tangent line, check the picture below, low and behold, Q = 0 when that happens.