$Let \ f=x^3+3xy^2+72x-15x^2-15y^2 $ Differentiating f partially w.r.t.x, $f_x=3x^2+3y^2+72-30x$ Differentiating f partially w.r.t.y, $f_y=6xy-30y$ Differentiating fx partially w.r.t.x, $f_{xx}=6x-30$ Differentiating fy partially w.r.t.y. $f_{yy}=6x-30$ and lastly, Differentiating fy partially w.r.t.x, $f_{xy}=6y$ $D=f_{xx}f_{yy}-f^2_{xy}= (6x-30)^2-(6y)^2$ For extreme values $f_x=0 $ and $f_y=0$ $3x^2+3y^2+72-30x=0 \ and \ 6xy-30y=0$ Solving them simultaneously, $6xy=30y, \ xy=5y, \ y=0, \ x=5$ $when \ y=0, \ 3x^2+3y^2+72-30x=0 \ gives, $ $3x^2-30+72=0, \ x=4,6$ $when \ x=5.3x^2+3y^2+72-30x=0 \ gives$ $75+3y^2+72-150=0 \to y=1, \ -1$ Hence the stationary values are, $(4,0)(6,0)(5,1)\ and \ (5,-1)$ $consider \ point \ (4,0),$ \(D=(6(4)-30)^2-(0)^3=36 \gt6\) Hence, (4,0) is an extreme and the value of f will be $4^3+0+72(4)-15(16)-0=112$ \(f_{xx}=6\times 4-30 = -6\lt0\) Hence f is maximum at (4,0) with max value=112. Consider, (6,0) \(D=(6(6)-30)^2-(0)^2=36\gt0\) Hence, (6,0) is an extreme and the value of f will be $6^3+0+72(6)-15(36)-0=108$ \(f_{xx}=6\times 6 -30 = 6 \gt0\) Hence f is minimum at (6,0) with max value =108. Cosider, (5,1), \(D=(6(5)-30)^2-(6)^2=-36\lt0\) So, (5,1) is not an extreme value. Cosider, (5,-1), \(D=(6(5)-30)^2-(-6)^2=-36\lt0\) So, (5,-1) is not an extreme value. Hence the extreme values of $x^3+3xy^2+72x-15x^2-15y^2, $ are 112 and 108 occurring at (4,0) and (6,0) respectively.