Discuss the maxima and minima of \(f(x,y)=x^3+3xy^2-15x^2-15y^2+72x\).

Answer: Maximum at \((4,0)\) with \(f=112\); minimum at \((6,0)\) with \(f=108\); \((5,\pm1)\) are saddle points (no extremum).

Stationary points: \(f_x=3x^2+3y^2-30x+72=0\Rightarrow x^2+y^2-10x+24=0\) ...(1); \(f_y=6xy-30y=0\Rightarrow 6y(x-5)=0\) ...(2).

From (2): \(y=0\) or \(x=5\).

• \(y=0\): (1) gives \(x^2-10x+24=0\Rightarrow x=4,6\). Points \((4,0),(6,0)\).
• \(x=5\): (1) gives \(25+y^2-50+24=0\Rightarrow y^2=1\Rightarrow y=\pm1\). Points \((5,1),(5,-1)\).

Second derivatives: \(r=f_{xx}=6x-30,\ s=f_{xy}=6y,\ t=f_{yy}=6x-30\).

Values: \(f_{max}=f(4,0)=64-240+288=112\); \(f_{min}=f(6,0)=216-540+432=108\).