Question

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

Answer 1

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

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.