Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 19 / a

Let $h=x^2+y^2+z-x$. Using Lagrange multipliers with $\nabla f=\lambda \nabla h$ givesThe third equation fixes $\lambda =1$. The first two then reduce to $2x-ay=1$ and $2y=ax$. If $a^2\ne 4$, their solution and the constraint giveAt $a=2$, the second equation requires $y=x$ and makes the first inconsistent; at $a=-2$, it requires $y=-x$ and gives the same inconsistency. Thus there is no constrained stationary point when $a=\pm 2$. The constraint gradient never vanishes because its vertical component is one, so the multiplier condition has not excluded any regular constrained stationary points.

Eliminate $z$ using the constraint: $z=x-x^2-y^2$. The restricted objective is the quadratic polynomial $F(x,y)=axy+x-x^2-y^2$, with constant Hessian matrixIts eigenvalues are $-2+a$ and $-2-a$. HenceFor $|a|<2$, both eigenvalues are negative and the quadratic polynomial is strictly concave, so the maximum is global on the entire constraint surface. For $|a|>2$, the opposite signs give increasing and decreasing directions. At $a=\pm 2$ there is no stationary point, as already found, rather than a degenerate extremum to classify. No parameter value produces a constrained local minimum.