Constrained Optimization Practice: Six Questions and Solutions

Write $y=6-x$. What values may $x$ take? Complete the square before deciding whether a maximum exists.

The feasible set is the whole line, so $x\in\mathbb R$. Substitution gives

Thus the unique global minimum is $24$ at $(2,4)$. As $|x|\to\infty$, $F(x)\to+\infty$, so there is no global maximum. The line is unbounded, but that does not prevent a minimum.

For $g=x+y=6$, $\nabla g=(1,1)\ne\mathbf0$. The equations

give $y=2x$, hence $(2,4)$, $\lambda=8$. Check the original constraint: $2+4=6$. Check the objective: $2(2)^2+4^2=24$. The completed square supplies the global proof.

Now $0\le x\le6$. The only interior stationary point remains $x=2$. At the endpoints, $f(0,6)=36$ and $f(6,0)=72$. Comparing these with $24$ gives the same unique minimum at $(2,4)$ and a global maximum of $72$ at $(6,0)$. The maximum exists because the endpoints restrict how far along the line you may go.

Use $(3,4)=\lambda(2x,2y)$. Can $\lambda$ be zero? Check that your points lie on the circle.

The circle is nonempty, closed and bounded, and $f$ is continuous. Both global extrema exist. The constraint gradient $(2x,2y)$ is nonzero everywhere on the circle, which has no endpoints.

The equations are

Here $\lambda=0$ is impossible because the first equation would give $3=0$. Thus $x=3/(2\lambda)$ and $y=4/(2\lambda)$. The constraint becomes $25/(4\lambda^2)=25$, so $\lambda=\pm1/2$.

The candidates are $(3,4)$, with value $25$, and $(-3,-4)$, with value $-25$. They give the global maximum and minimum respectively. Every possible global extremum must satisfy these candidate equations, and every solution has been checked.

Independent check: Cauchy–Schwarz gives $$\begin{gathered}|3x+4y|\\\le\sqrt{3^2+4^2}\sqrt{x^2+y^2}\\=25\end{gathered}$$, with equality at exactly these aligned or oppositely aligned points.

For every feasible point, $|3x+4y|<25$; neither boundary point is allowed. For $0<t<1$, the points $(3t,4t)$ give $f=25t\to25$ as $t\to1^-$, and $(-3t,-4t)$ give $f=-25t\to-25$. Therefore the supremum is $25$ and the infimum is $-25$, but neither is attained. There is no global maximum or minimum. A supremum is the least upper bound; an infimum is the greatest lower bound. They need not be values at feasible points.

Use $y=2\lambda x$ and $x=8\lambda y$. Check zero-coordinate cases before cancelling. Relate $x^2$ and $y^2$ without choosing signs too early.

The ellipse is compact, and $\nabla g=(2x,8y)$ is never zero on it. Both global extrema exist and must appear among the regular candidates.

If $x=0$, the first multiplier equation forces $y=0$, which is infeasible. Similarly $y=0$ cannot be a candidate. Multiplying the first equation by $y$ and the second by $x$ yields

so $x^2=4y^2$. The constraint then gives $8y^2=4$, or $y^2=1/2$, and $x^2=2$.

The global maximum is $1$ at $(\sqrt2,1/\sqrt2)$ and $(-\sqrt2,-1/\sqrt2)$. The global minimum is $-1$ at $(\sqrt2,-1/\sqrt2)$ and $(-\sqrt2,1/\sqrt2)$. All four satisfy the constraint and multiplier equations; same signs give $\lambda=1/4$, opposite signs give $\lambda=-1/4$.

Independent check: $(x-2y)^2\ge0$ and $(x+2y)^2\ge0$, together with the constraint, imply $-1\le xy\le1$. The listed points attain the two bounds.

The constraint itself gives $h=4$ at every feasible point. Thus the global maximum and minimum are both $4$, each attained everywhere on the ellipse. They are not strict extrema. With $g=x^2+4y^2$, the equations $\nabla h=\lambda\nabla g$ hold at every point for $\lambda=1$. A candidate set need not be a finite list.

Inside the disk, use $\nabla f=\mathbf0$. On the circle, use the constraint to rewrite $f$, or solve the multiplier equations while keeping zero-coordinate cases.

Continuity on the closed disk guarantees both global extrema. In the interior, $\nabla f=(2x,4y)=\mathbf0$ gives the feasible point $(0,0)$, with value $0$.

On the boundary, $x^2+y^2=4$, so $f=4+y^2$. Its boundary minimum is $4$ at $(\pm2,0)$, and its boundary maximum is $8$ at $(0,\pm2)$.

For completeness, the boundary multiplier equations are

If both coordinates were nonzero, the first equation would give $\lambda=1$ and the second $\lambda=2$, impossible. If $y=0$, the candidates are $(\pm2,0)$; if $x=0$, they are $(0,\pm2)$. Both cannot be zero on the circle. No boundary cases are missing.

Comparing the five points gives the unique global minimum $0$ at $(0,0)$ and global maximum $8$ at both $(0,2)$ and $(0,-2)$. A direct bound checks the result: $$\begin{aligned}0\le x^2+2y^2&\le2(x^2+y^2)\\&\le8\end{aligned}$$, with the stated equality cases. The boundary minimum $4$ is not the minimum on the disk.

The origin is now infeasible. The global minimum is $4$ at $(\pm2,0)$; the global maximum remains $8$ at $(0,\pm2)$. The objective did not change, but the feasible set did.

Which set does the squared equation describe? Compute its gradient on that set. Can the same set be represented by a regular equation?

The feasible set is exactly the unit circle. If $G=(x^2+y^2-1)^2$, then

at every feasible point. Since $\nabla f=(1,1)$, the equations $\nabla f=\lambda\nabla G$ have no solutions. The regularity hypothesis has failed everywhere; those equations are not a necessary condition here.

Use the equivalent regular constraint $g=x^2+y^2=1$. Now $(1,1)=2\lambda(x,y)$, so $\lambda\ne0$ and $x=y$. The constraint gives $x=y=\pm1/\sqrt2$. The continuous objective on the compact circle therefore has maximum $\sqrt2$ at $(1/\sqrt2,1/\sqrt2)$ and minimum $-\sqrt2$ at $(-1/\sqrt2,-1/\sqrt2)$. The bound $|x+y|\le\sqrt2$ provides an independent check.

The feasible parametrization $(x,y)=(t^2,t^3)$, $t\in\mathbb R$, covers the whole cusp. Indeed, the constraint forces $x\ge0$; for $x>0$, setting $t=y/x$ gives $t^2=x$ and $t^3=y$, while the origin corresponds to $t=0$.

Then $h=t^4+t^6\ge0$, with equality only at $t=0$. The unique global minimum is $0$ at $(0,0)$; there is no global maximum because $h\to+\infty$ as $|t|\to\infty$. For $g=y^2-x^3$, $\nabla g=(-3x^2,2y)$ vanishes at the origin, so it deserves a direct check even though it happens also to satisfy the multiplier equation for this objective.

Write $F(x)=f(x,2)$. A direct sign comparison can settle the problem even when the second-derivative test cannot.

The regular constraint $g=y=2$ has gradient $(0,1)$. Lagrange gives

so the candidate is $(1,2)$, with $\lambda=4$. Substitution gives

Thus $F''(1)=0$, and the second-derivative test is inconclusive. However, $F(x)\ge4$, with equality only at $x=1$. The unique global minimum is $4$ at $(1,2)$. There is no global maximum because the fourth power is unbounded above. The inequality supplies the missing classification.

The candidate is again $(1,2)$: $3(x-1)^2=0$, $2y=\lambda$, $y=2$. But the reduced function is $H(x)=(x-1)^3+4$. For every $\varepsilon>0$,

Hence the candidate is neither a local minimum nor a local maximum along the line. As $x\to\pm\infty$, the objective is unbounded in both directions, so there is no global extremum either. Again the second derivative at the candidate is zero; the sign comparison resolves what that test leaves open.