Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 19

The second partial derivatives are $f_{xx}=4$, $f_{xy}=f_{yx}=3+2y$ and $f_{yy}=2+2x+30y$. Evaluating at the origin gives the Hessian matrix

The determinant of the Hessian matrix is $4\cdot 2-3^2=-1<0$. Hence it has one positive and one negative eigenvalue (explicitly $3\pm \sqrt{10}$), so its quadratic form is indefinite. The cubic terms are of higher order and cannot remove these opposite local directions of change. ThereforeFor example the quadratic term is positive along $y=0$ but negative along $y=-3x/2$, confirming the saddle point of a scalar function classification.

For a right prism with equilateral triangular ends, the triangle area is $\sqrt{3}{\alpha }^2/4$. Thus its volume and external area areApply the Lagrange multiplier method to $V-\lambda (A-A_0)$ with positive lengths. The stationarity equations areThe second gives $\lambda =\sqrt{3}\alpha /12$. Substitution in the first then gives $6\beta =\sqrt{3}\alpha +3\beta$, or $\beta =\alpha /\sqrt{3}$. The area constraint becomesConsequentlyThe resulting volume is $V_{\max }=3^{3/4}/4{\mathrm{m}}^3$. As a check on the maximum, eliminating $\beta$ gives $V(\alpha )=\sqrt{3}{A}_0\alpha /12-{\alpha }^3/8$, whose second derivative is negative for every admissible positive $\alpha$; volume also tends to zero at both degenerate boundary shapes.