Budget-constrained quadratic models

An equality such as $x+y=1$ describes which choices are allowed. A quadratic expression describes how those choices are compared. These are different parts of an optimization problem: solving derivative equations is useful only after the feasible set and the objective have both been specified.

We study a hypothetical mathematical model. The symbols $x$ and $y$ are real weights and the objective is a stated quadratic criterion. No empirical interpretation of its coefficients is needed for the calculations below.

1. State the feasible set before differentiating

Consider

$$V(x,y)=4x^2+2xy+9y^2, \qquad x+y=1.$$

With only the equality constraint, the feasible set is the entire line

$$F_{\mathbb R}=\{(x,1-x):x\in\mathbb R\}.$$

For example, $(2,-1)$ satisfies the equality. If negative weights are forbidden, the problem has the additional assumptions $x\ge0$ and $y\ge0$. Since $y=1-x$, its feasible set becomes the closed segment

$$F_+=\{(x,1-x):0\le x\le1\}.$$

A minimum over the line and a minimum over this segment are different claims. An equality-constrained candidate may fail an additional inequality constraint; we must check it rather than assume feasibility.

2. Derive the candidate by eliminating one variable

Every feasible point has exactly one representation $(x,1-x)$. Substitution therefore loses no feasible choices:

$$\begin{aligned} g(x)=V(x,1-x) &=4x^2+2x(1-x)+9(1-x)^2\\ &=4x^2+2x-2x^2+9-18x+9x^2\\ &=11x^2-16x+9. \end{aligned}$$

Here $g'(x)=22x-16$. Its only stationary point is $x_*=8/11$, so $y_*=1-8/11=3/11$. Both weights are positive and sum to one.

The derivative locates a candidate. The next step establishes why it is a global minimum and why no other feasible point attains the same value.

3. Prove global optimality with an exact square

Completing the square gives the identity

$$g(x)=\frac{35}{11}+11\left(x-\frac8{11}\right)^2.$$

The coefficient 11 is positive. Consequently $g(x)\ge35/11$ for every real $x$, with equality exactly when $x=8/11$. Thus $(8/11,3/11)$ is the unique global minimizer over $F_{\mathbb R}$, and the minimum value is $35/11$.

Because this point also belongs to $F_+$, the same point and value solve the nonnegative-weight problem. The endpoints are a useful independent check:

$$V(0,1)=9,\qquad V(1,0)=4,$$

both larger than $35/11$. The square identity is stronger than checking only endpoints: it compares the candidate with every point of the entire feasible line.

For a more restrictive interval $a\le x\le b$, the same identity says to choose the point of $[a,b]$ nearest to $8/11$, then set $y=1-x$. If the interval excludes $8/11$, the minimizer moves to a boundary. This conclusion follows from the square; it does not require pretending a boundary point is stationary.

4. Reconcile the multiplier and matrix calculations

Using the convention

$$L(x,y,\lambda)=V(x,y)-\lambda(x+y-1),$$

the multiplier equations are

$$8x+2y=\lambda,\qquad 2x+18y=\lambda,\qquad x+y=1.$$

Subtracting the first two equations gives $6x-16y=0$. Hence $x=(8/3)y$, and the budget gives $(11/3)y=1$. We recover $y=3/11$, $x=8/11$ and $\lambda=70/11$. These equations agree with substitution. Their agreement does not replace the preceding global comparison.

For $w=(x,y)^T$ the same criterion is

$$V=w^T\Sigma w, \qquad \Sigma=\begin{pmatrix}4&1\\1&9\end{pmatrix}.$$

The two off-diagonal entries contribute $xy+yx=2xy$. Because $\Sigma$ is symmetric, $\nabla V=2\Sigma w=(8x+2y,2x+18y)^T$, explaining the multiplier equations in matrix notation. Calling this quadratic a variance model would require separate probabilistic assumptions about $\Sigma$; algebra alone supplies no such interpretation.

Exercises

P08. Add the constraint $0\le x\le1/2$ to the stated model. Find the minimizing weights and the minimum value. Explain why the equality-only stationary point is no longer feasible.

P08a. Expand $w^T\Sigma w$ for the displayed matrix. Explain why putting 2 in both off-diagonal entries would give a different objective.

P08b. Instead require $3/4\le x\le1$, with $y=1-x$. Find the minimizing weights and value by using distance from $8/11$.

P08c. Replace the objective by $-V$ but keep the entire feasible line $F_{\mathbb R}$. Does the same stationary point minimize the new objective? Explain both what happens there and whether a global minimum exists.

Full solutions

P08. Since $8/11>1/2$, the equality-only candidate is excluded. On $[0,1/2]$, the point nearest to $8/11$ is $1/2$. Therefore $(x,y)=(1/2,1/2)$, and direct substitution gives

$$V(1/2,1/2)=\frac44+\frac24+\frac94=\frac{15}{4}.$$

This boundary point need not satisfy $g'(x)=0$; indeed $g'(1/2)=-5$.

P08a. Multiplication gives $4x^2+xy+yx+9y^2=4x^2+2xy+9y^2$. With both off-diagonal entries equal to 2, the cross terms would sum to $4xy$, changing the criterion.

P08b. Since $8/11<3/4$, the nearest feasible point is $x=3/4$. Thus $y=1/4$, and

$$V(3/4,1/4)=\frac{36+6+9}{16}=\frac{51}{16}.$$

P08c. Along the line, $-V=-35/11-11(x-8/11)^2$. The same point is now the unique global maximum. As $|x|\to\infty$, the objective tends to $-\infty$, so a global minimum does not exist on that line. Stationary equations alone therefore cannot identify the type of optimum.