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Checking Lagrange Candidates with Parametrizations

Use a parametrization to verify which Lagrange candidates are genuine constrained extrema and to confirm their values.

Intermediate 10 min read

What You'll Learn

  • Parameterize a circle or ellipse so that the parametrization covers the entire feasible curve.
  • Differentiate the resulting one-variable objective to find constrained candidates.
  • Relate the one-variable derivative condition to the Lagrange multiplier condition.
  • Compare candidate points and objective values from both methods.
  • Avoid losing candidates by dividing by a factor or restricting the parameter interval.

Why Check Lagrange Candidates with a Parametrization?

A Lagrange multiplier calculation turns a constrained optimization problem into a system of equations. Solving that system produces candidate points, but the algebra can be difficult to interpret, and a candidate list is only useful if it contains all the relevant points. When the feasible set is a curve that can be parametrized simply, there is an independent check: describe every feasible point using one parameter, then optimize the resulting one-variable function.

This method does more than confirm arithmetic. It shows directly how the objective changes as we move along the constraint, helps identify missed candidates, and makes it possible to compare the values at all stationary points. The parametrization must cover the whole feasible curve; checking only part of a circle, for example, can miss a minimum.

Definition: A parametrization of a feasible curve is a function \(\mathbf{r}(t)=(x(t),y(t))\) whose values lie on the curve. It covers the curve if every feasible point is \(\mathbf{r}(t)\) for at least one allowed parameter value \(t\). For an objective \(f(x,y)\), the corresponding one-variable objective is \(h(t)=f(x(t),y(t))\).

For a circle of radius \(R>0\) centered at the origin, a full parametrization is \(\mathbf{r}(t)=(R\cos t,R\sin t)\), with \(0\leq t\leq 2\pi\). For an ellipse described by \(x^2/a^2+y^2/b^2=1\), where \(a,b>0\), a full parametrization is \(\mathbf{r}(t)=(a\cos t,b\sin t)\) over the same interval. The endpoints represent the same point, which is harmless: they ensure the entire closed curve is covered.

How the Two Candidate Conditions Agree

Suppose a smooth closed curve is described by \(g(x,y)=c\), with \(\nabla g\neq\mathbf{0}\) on the curve. Assume \(\mathbf{r}(t)=(x(t),y(t))\) parametrizes the curve smoothly and \(\mathbf{r}'(t)\neq\mathbf{0}\). By the chain rule, the derivative of the objective along the curve is

$$ h'(t)=\nabla f(\mathbf{r}(t))\mathbin{\cdot}\mathbf{r}'(t). $$

Also, differentiating the constraint identity \(g(\mathbf{r}(t))=c\) gives \(\nabla g(\mathbf{r}(t))\mathbin{\cdot}\mathbf{r}'(t)=0\). Thus the velocity \(\mathbf{r}'(t)\) is tangent to the constraint curve, while the nonzero gradient \(\nabla g\) is perpendicular to that tangent. The condition \(h'(t)=0\) says that \(\nabla f\) is also perpendicular to the tangent. In the plane, the vectors perpendicular to this nonzero tangent form one line, so \(\nabla f\) is a scalar multiple of \(\nabla g\). That is the Lagrange multiplier condition.

What this establishes: At a regular point of a smoothly parametrized constraint curve, a stationary point of the objective along the curve satisfies the Lagrange multiplier equations. Conversely, a point satisfying those equations has zero derivative along the curve. The parametrization must cover the feasible curve, and the regularity conditions matter: at a singular point, the multiplier equations may not apply.

To find absolute extrema, stationary points alone are not enough: evaluate the objective at every candidate and compare the values. On a closed, bounded curve, a continuous objective attains both an absolute maximum and an absolute minimum. A full parametrization lets us use the corresponding one-variable function to locate and compare them.

Worked Example: A Linear Objective on a Circle

Worked Example: A Linear Objective on a Circle

Find the maximum and minimum of \(f(x,y)=3x+4y\) subject to \(x^2+y^2=4\), and compare the Lagrange and parametrization calculations.

First use Lagrange multipliers with \(g(x,y)=x^2+y^2\). The equations \(\nabla f=\lambda\nabla g\) are \(3=2\lambda x\) and \(4=2\lambda y\), together with \(x^2+y^2=4\). Since the first two equations require \(\lambda\neq 0\), they give \(x=3/(2\lambda)\) and \(y=2/\lambda\). Substitution into the constraint yields

$$ \frac{9}{4\lambda^2}+\frac{4}{\lambda^2}=4 \quad\Longrightarrow\quad \frac{25}{4\lambda^2}=4 \quad\Longrightarrow\quad \lambda^2=\frac{25}{16}. $$

Thus \(\lambda=5/4\) or \(\lambda=-5/4\). The corresponding points are \((6/5,8/5)\) and \((-6/5,-8/5)\). Their objective values are \(18/5+32/5=10\) and \(-18/5-32/5=-10\).

Now parametrize the entire circle by \(x=2\cos t\), \(y=2\sin t\), for \(0\leq t\leq 2\pi\). The objective becomes \(h(t)=6\cos t+8\sin t\), so

$$ h'(t)=-6\sin t+8\cos t. $$

At a stationary point, \(8\cos t=6\sin t\). Together with \(\cos^2t+\sin^2t=1\), this gives \((\cos t,\sin t)=(3/5,4/5)\) or \((-3/5,-4/5)\). These yield the same two points found by the multiplier calculation. Their values are \(6(3/5)+8(4/5)=10\) and \(-6(3/5)-8(4/5)=-10\). Since these are the only stationary values on the closed curve, they are the absolute maximum and minimum. Substitution also checks feasibility: each point has \(x^2+y^2=36/25+64/25=4\).

Worked Example: A Product Objective on a Circle

Worked Example: A Product Objective on a Circle

Find the absolute extrema of \(f(x,y)=x^2y\) subject to \(x^2+y^2=1\), and use the parametrization to check the full candidate set.

For the multiplier calculation, \(\nabla f=(2xy,x^2)\) and \(\nabla g=(2x,2y)\). The equations are \(2xy=2\lambda x\), \(x^2=2\lambda y\), and \(x^2+y^2=1\). If \(x=0\), the constraint gives \(y=1\) or \(y=-1\), and both points satisfy the multiplier equations with \(\lambda=0\). If \(x\neq 0\), the first equation gives \(\lambda=y\). The second then gives \(x^2=2y^2\). Substituting into the constraint gives \(3y^2=1\), so \(y=\pm 1/\sqrt{3}\) and \(x=\pm\sqrt{2/3}\). Each sign of \(x\) is possible. At these four points, \(f=x^2y=(2/3)y\), giving values \(2/(3\sqrt{3})\) when \(y=1/\sqrt{3}\) and \(-2/(3\sqrt{3})\) when \(y=-1/\sqrt{3}\). At \((0,1)\) and \((0,-1)\), the value is \(0\).

For an independent check, set \(x=\cos t\), \(y=\sin t\), over \(0\leq t\leq 2\pi\). Then

$$ h(t)=\cos^2t\sin t, \qquad h'(t)=\cos t\bigl(1-3\sin^2t\bigr). $$

To solve \(h'(t)=0\), set each factor equal to zero. The first gives \(\cos t=0\), or \((x,y)=(0,1)\) and \((0,-1)\), where \(h=0\). The second gives \(\sin^2t=1/3\), so \(y=\pm1/\sqrt{3}\) and \(x=\pm\sqrt{2/3}\). The resulting values agree exactly with the multiplier calculation. Comparing all values, the absolute maximum is \(2/(3\sqrt{3})\), and the absolute minimum is \(-2/(3\sqrt{3})\). At the four nonzero candidates, the constraint checks as \(2/3+1/3=1\); the two points with \(x=0\) also satisfy it.

Worked Example: A Linear Objective on an Ellipse

Worked Example: A Linear Objective on an Ellipse

Find the absolute extrema of \(f(x,y)=x+y\) subject to \(x^2/4+y^2=1\), and compare both methods.

Set \(g(x,y)=x^2/4+y^2\), so \(\nabla g=(x/2,2y)\). The multiplier equations \((1,1)=\lambda(x/2,2y)\) give \(x=2/\lambda\) and \(y=1/(2\lambda)\). Substitution in the constraint gives

$$ \frac{(2/\lambda)^2}{4}+\left(\frac{1}{2\lambda}\right)^2 = \frac{1}{\lambda^2}+\frac{1}{4\lambda^2} = 1, \qquad \lambda^2=\frac54. $$

Thus \(\lambda=\pm\sqrt{5}/2\). The corresponding points are \((4/\sqrt{5},1/\sqrt{5})\) and \((-4/\sqrt{5},-1/\sqrt{5})\), with objective values \(\sqrt{5}\) and \(-\sqrt{5}\). Each point is feasible because \(x^2/4=4/5\) and \(y^2=1/5\).

Parametrize the full ellipse by \(x=2\cos t\), \(y=\sin t\). Then \(h(t)=2\cos t+\sin t\), and \(h'(t)=-2\sin t+\cos t\). A stationary point satisfies \(\cos t=2\sin t\). With \(\cos^2t+\sin^2t=1\), this gives \((\cos t,\sin t)=(2/\sqrt{5},1/\sqrt{5})\) or \((-2/\sqrt{5},-1/\sqrt{5})\). The resulting points and values are exactly those obtained from the multiplier equations. As the ellipse is closed and bounded, comparing these values gives the absolute maximum \(\sqrt{5}\) and minimum \(-\sqrt{5}\).

Common Pitfall: Dividing Away a Candidate

In the product example, the derivative factors as \(h'(t)=\cos t(1-3\sin^2t)\). A tempting but incorrect shortcut is to divide by \(\cos t\), then solve only \(1-3\sin^2t=0\). That division is valid only when \(\cos t\neq 0\). It discards the cases \(\cos t=0\), which give the stationary points \((0,1)\) and \((0,-1)\). Although these particular points do not give the absolute extrema, they are part of the candidate set and have value \(0\).

Correction: When a derivative is written as a product, set each factor equal to zero and check its solutions. Do not divide by a factor unless you separately account for the case in which that factor is zero.

A second check is to make sure the parameter interval covers the whole feasible curve. Restricting the circle parameter in the product example to \(0\leq t\leq\pi\) would include \(y\geq 0\), but not the lower half of the circle. It would miss the negative objective values and therefore the absolute minimum. A correct candidate check needs both a complete parametrization and a complete comparison of the resulting values.

When This Check Is Useful

Parametrizing a feasible curve is especially useful when the constraint is a circle, ellipse, or another curve with a simple full parametrization. It can reveal a missed case in multiplier algebra, confirm that the candidate points satisfy the constraint, and show which candidates give maxima or minima. It also gives a direct picture of the objective as the parameter moves around the curve.

The two methods answer related but distinct questions. The multiplier equations produce points where the objective has no first-order change in a tangent direction. The parametrization turns that tangent-direction test into the ordinary condition \(h'(t)=0\). In either method, the absolute extrema are identified only after all candidates have been found and their objective values compared.

Check Your Understanding

Use a full parametrization where appropriate, and explain how you know you have not omitted candidates.

  1. For \(f(x,y)=x-y\) on \(x^2+y^2=1\), write the one-variable objective and its derivative.
  2. For \(f(x,y)=xy\) on \(x^2+y^2=1\), what factors appear in the derivative after setting \(x=\cos t\) and \(y=\sin t\)? Why should each factor be considered?
  3. State the regularity conditions used to relate a zero derivative along a parametrized curve to the Lagrange multiplier condition.
  4. Why is finding stationary points not, by itself, enough to identify an absolute maximum or minimum?
  5. What portion of the ellipse \(x^2/4+y^2=1\) is covered when \(0\leq t\leq\pi\), \(x=2\cos t\), and \(y=\sin t\)?

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