Show that the minimum value of \(u=xy+a^3\left(\dfrac1x+\dfrac1y\right)\) is \(3a^2\). (asked in MU June 2025 KT)

Answer: The only stationary point is \((a,a)\); there \(rt-s^2=3>0\) and \(r=2>0\), so \(u_{min}=3a^2\) (taking \(a>0\)).

\(u_x=y-\dfrac{a^3}{x^2}=0\Rightarrow x^2y=a^3\); \(u_y=x-\dfrac{a^3}{y^2}=0\Rightarrow xy^2=a^3\).

Dividing: \(\dfrac{x^2y}{xy^2}=1\Rightarrow x=y\). Then \(x^3=a^3\Rightarrow x=y=a\). Stationary point \((a,a)\).

\(r=u_{xx}=\dfrac{2a^3}{x^3},\ s=u_{xy}=1,\ t=u_{yy}=\dfrac{2a^3}{y^3}\). At \((a,a)\): \(r=2,\ s=1,\ t=2\).

\(rt-s^2=4-1=3>0\) and \(r=2>0\) ⇒ \(u\) is minimum at \((a,a)\).

\[u_{min}=a\cdot a+a^3\left(\frac1a+\frac1a\right)=a^2+2a^2=3a^2.\]