Critical Points of an Unconstrained Cubic Surface: 23-Chem-B3 Simulation, Modelling, and Optimization, December 2014, Q2

Question 2: Critical Points of an Unconstrained Cubic Surface (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The unconstrained objective $f(x,y) = x^3 - 7xy + 6y^3$ to be maximized over all real $(x,y)$.

Find. (a) whether a maximum exists (and its $x$-value if so); (b) all other critical points and their nature (max / min / saddle).

Approach. Locate the stationary points from $\nabla f = 0$, then classify each with the second-order (Hessian) test $D = f_{{xx}}f_{{yy}} - f_{{xy}}^2$.

First-order conditions. The gradient must vanish: $$f_x = 3x^2 - 7y = 0,\qquad f_y = -7x + 18y^2 = 0.$$
Solve the system. From the first, $y = \tfrac{3}{7}x^2$. Substituting into the second, $$-7x + 18\left(\tfrac{3}{7}x^2\right)^2 = -7x + \tfrac{162}{49}x^4 = x\left(\tfrac{162}{49}x^3 - 7\right) = 0,$$ so $x = 0$ or $x^3 = \tfrac{343}{162}$, i.e. $x = \boxed{1.284}$ with $y = \tfrac{3}{7}x^2 = 0.707$. The two stationary points are $(0,0)$ and $(1.284,\ 0.707)$.
Hessian. $f_{{xx}} = 6x,\; f_{{yy}} = 36y,\; f_{{xy}} = -7$, so the discriminant is $$D(x,y) = f_{{xx}}f_{{yy}} - f_{{xy}}^2 = 216\,xy - 49.$$
Classify $(0,0)$. $D(0,0) = -49 \lt 0$, so the origin is a saddle point (not an extremum).
Classify $(1.284,\,0.707)$. $D = 216(1.284)(0.707) - 49 = +147 \gt 0$ and $f_{{xx}} = 6(1.284) = 7.7 \gt 0$, so this point is a local minimum.
Answer part (a). Neither stationary point is a maximum, and $f$ is unbounded above (as $x\to+\infty$ with $y$ fixed, $x^3\to+\infty$; likewise the $6y^3$ term). Hence $$\boxed{\text{no maximum exists — the function is unbounded above.}}$$

Point	$D = 216xy-49$	Nature
$(0,\ 0)$	$-49\ (\lt 0)$	Saddle point
$(1.284,\ 0.707)$	$+147\ (\gt 0),\ f_{{xx}}\gt0$	Local minimum
Global maximum	—	Does not exist (unbounded)

Concept & Theory

Topic. Second-order optimality conditions for unconstrained multivariable optimization.

Key relations. Stationarity $\nabla f = 0$; Hessian test with $D = f_{{xx}}f_{{yy}} - f_{{xy}}^2$: $D\gt0,\,f_{{xx}}\gt0\Rightarrow$ min; $D\gt0,\,f_{{xx}}\lt0\Rightarrow$ max; $D\lt0\Rightarrow$ saddle; $D=0\Rightarrow$ inconclusive.

Why this works. A polynomial with odd-degree leading terms in each variable can never attain a global maximum — it runs off to $+\infty$. The interior stationary points are therefore at best local, and the Hessian tells them apart. Recognizing "no global optimum" is itself a modelling result: an unconstrained maximization of this $f$ is ill-posed and would need bounds or constraints to be meaningful.

Common pitfall. Concluding a "maximum" was found merely because the gradient vanished. A zero gradient only signals a stationary point; the Hessian (or global growth) must confirm the type.

Source. Rao, Engineering Optimization §2.3 (unconstrained optimization, Hessian test); Edgar, Himmelblau & Lasdon, ch. 4.

Question 2 of 7: Critical Points of an Unconstrained Cubic Surface

Question 2: Critical Points of an Unconstrained Cubic Surface (20 marks)