= Solution
Write $\gamma'=d\gamma/dT<0$. The <Maxwell relation> gives
$$
\left(\frac{\partial S}{\partial A}\right)_T=-\gamma'>0.
$$
Thus \b[isothermal stretching increases the <entropy>]. For an infinitesimal increase, $dS=-\gamma'\,dA$; if the <derivative> is independent of <area>, this integrates to $\Delta S=-\gamma'\Delta A$ at fixed <temperature>. Mechanical <energy> used to create extra surface is not, in general, identical to the increase of <internal energy> at fixed <temperature>, because the stretching can also absorb <heat>.
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