Q.6 (b) Show that the minimum value of u=xy+a³(1/x+1/y) is 3a² .

u_x=y-a³/x², u_y=x-a³/y² . Stationary points: x²y=a³,\ xy²=a³ . Dividing, x=y ; then x³=a³⇒ x=y=a . Point (a,a) . r=u_xx=2a³/x³,\ s=u_xy=1,\ t=u_yy=2a³/y³ . At (a,a) : r=2,\ s=1,\ t=2 . rt-s²=4-1=3>0 and r=2>0 → minimum at (a,a) . u_min=a· a+a³(1/a+1/a)=a²+2a² → minimum value =3a² (proved).

Show that the minimum value of \(u=xy+a^3\left(\frac1x+\frac1y\right)\) is \(3a^2\).

Q.6 (b) Show that the minimum value of \(u=xy+a^3\left(\frac1x+\frac1y\right)\) is \(3a^2\).

Doubtly Asked 25 Oct 2026 mu engineering Save WhatsApp Report

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\(u_x=y-\frac{a^3}{x^2},\quad u_y=x-\frac{a^3}{y^2}\).

Stationary points: \(x^2y=a^3,\ xy^2=a^3\). Dividing, \(x=y\); then \(x^3=a^3\Rightarrow x=y=a\). Point \((a,a)\).

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

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

\(u_{min}=a\cdot a+a^3\left(\frac1a+\frac1a\right)=a^2+2a^2\) → minimum value \(=3a^2\) (proved).

Doubtly AI · 25 Oct 2026 AI-verified · Oct 2026 Report

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Show that the minimum value of \(u=xy+a^3\left(\frac1x+\frac1y\right)\) is \(3a^2\).

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