A calculus problem by Steven Chase

These are not the shortest solutions I found, but they seemed more insightful. As is common for polar representation of complex numbers, assume $A \ge 0$ and $B \ge 0$ . [There is a second solution if one of these is negative, but it is equivalent to the first solution except that the sought-after angle is supplementary to $\theta$ of the first solution.] Noting that $|v_1| = A$ and $|v_2| = B$ , the optimization problem is: $\begin{aligned} \min \; & |v_1|^2 + |v_2|^2 \\ \textrm{s.t. } \; \qquad &\\ &|v_1 + v_2| = 3,\\ &|v_1| + |v_2| = 5,\\ &\textrm{Im}(v_1) = 0,\\ &v_1, v_2 \in \mathbb{C}. \\ \end{aligned}$

[Approach 1] Geometrically, this can be viewed as finding a circle with radius $|v_2| = B$ centered on $(|v_1|,0) = (A,0)$ that (i) intersects a circle of radius 3 centered on the origin and (ii) intersects line $|v_1| + |v_2| = 5$ so that the distance from a point of intersection with the line and the origin is minimum - see the left figure . The line's point of closest approach to the origin occurs at $A=B \Rightarrow A = B = \frac{5}{2}$ . It remains to verify that $\theta$ is feasible at this point. Using $9 = |v_1 + v_2|^2 = A^2 + B^2 +2 A B \cos \theta = 2B^2(1+ \cos \theta ) \Rightarrow \cos \theta = -1+ \frac{18}{25} \in [-1,1]$ . So, $\boxed{\theta = \cos^{-1}(\frac{18}{25} -1) \approx 106.260^\circ }$ . [Approach 2] Employ Lagrange multipliers for the equality constraints. In terms of $A$ , $B$ , and $\theta$ , the optimization problem is: $\begin{aligned} \min \; & (A^2 + B^2) \\ \textrm{s.t. } \; \qquad &\\ &A^2 + B^2 +2 A B \cos\theta - 9 = 0,\\ &A + B - 5 = 0,\\ &A \ge 0, B \ge 0, \theta \in [0,2\pi] \\ \end{aligned}$

[Technically, the inequality constraints for the feasible region need to be treated as well, but if solution points are in the interior of it, the associated multipliers will be zero. For brevity, we assume solution points interior and just sketch the approach.] Introduce Lagrange multipliers $\lambda_1$ and $\lambda_2 \in \mathbb{R}$ for the two equality constraints and seek a stationary point in the Lagrangian $\mathscr{L}(A, B, \theta, \lambda_1, \lambda_2) = (A^2 + B^2) - \lambda_1(A^2 + B^2 +2 A B \cos\theta - 9) - \lambda_2(A + B - 5).$ Stationary points are found by solving the system of equations $\boldsymbol{\nabla} \mathscr{L} = \boldsymbol{0}$ for $(A, B, \theta, \lambda_1, \lambda_2)$ :

$\begin{aligned} (1) \; \dfrac{\partial \mathscr{L}}{\partial A} &= 2A - \lambda_2 - \lambda_1(2A + 2B\cos \theta) = 0 \\ (2) \; \dfrac{\partial \mathscr{L}}{\partial B} &= 2B - \lambda_2 - \lambda_1(2B + 2A\cos \theta) = 0 \\ (3) \; \dfrac{\partial \mathscr{L}}{\partial \theta} &= 2 \lambda_1 AB \sin \theta = 0 \\ (4) \; \dfrac{\partial \mathscr{L}}{\partial \lambda_1} &= A^2 + B^2 +2 A B \cos\theta - 9 = 0 \\ (5) \; \dfrac{\partial \mathscr{L}}{\partial \lambda_2} &= A + B - 5 = 0 \\ \end{aligned}$

Equation (3) has 4 possible cases that need to be considered: $\lambda_1 = 0, A = 0, B = 0,$ and $\sin \theta = 0.$

$\boldsymbol{\lambda_1 = 0:}$ (1) & (2) give $2A = \lambda_2 = 2B \Rightarrow A = B.$ So (5) $\Rightarrow A = B = \frac{5}{2}.$ So (4) $\Rightarrow 2B^2(1+ \cos \theta ) = 9 \Rightarrow$ Stationary point: $\color{#20A900}A=B= \frac{5}{2}, \theta = \cos^{-1}(\frac{18}{25} -1), \lambda_1=0, \lambda_2=5$ .

$\boldsymbol{A = 0:}$ (5) gives $B=5$ . But (4) $\Rightarrow 0 + B^2 + 0 - 9 = 0 \Rightarrow B = \pm 3$ , a contradiction. This case is infeasible .

$\boldsymbol{B = 0:}$ Same as $A = 0$ case: a contradiction. This case is infeasible .

$\boldsymbol{\sin \theta = 0:}$ Then $\cos \theta = \pm 1$ . Suppose $\cos \theta = +1$ . Then (4) $\Rightarrow (A+B)^2 = 9 \Rightarrow A + B = \pm 3$ , which contradicts (5), $A+B = 5$ . So, only possibility for this case is $\cos \theta = -1$ i.e., $\theta = \pi$ . Then (4) $\Rightarrow (A - B)^2 = 9 \Rightarrow A - B = \pm3$ . Combined with (5) $\Rightarrow$ two possibilities: $A=4, B=1$ or $A=1, B=4$ . It can be checked for both of these cases that $\lambda_1= \frac{1}{2}$ and $\lambda_2=5$ . So, two Stationary points: $\color{#D61F06} A=4, B=1, \theta = \pi, \lambda_1= \frac{1}{2}, \lambda_2=5$ and $\color{#EC7300} A=1, B=4, \theta = \pi, \lambda_1= \frac{1}{2}, \lambda_2=5$ .

All three of these stationary points occur in the interior of the feasible region (in agreement with the assumption). They need to be checked for optimality by inserting them in $\mathscr{L}$ or, more directly, into $A^2 + B^2$ . Doing so yields:

$\color{#D61F06} A=4, B=1, \theta = \pi, \lambda_1= \frac{1}{2}, \lambda_2=5$ or $\color{#EC7300} A=1, B=4, \theta = \pi, \lambda_1= \frac{1}{2}, \lambda_2=5$ $\Rightarrow A^2 + B^2 = 17$ ( red, orange solns in right fig ).

$\color{#20A900}\boxed{A=B= \frac{5}{2}, \theta = \cos^{-1}(\frac{18}{25} -1), \lambda_1=0, \lambda_2=5}$ $\Rightarrow A^2 + B^2 = \dfrac{25}{2}$ , which is minimal ( green soln in left fig ).

$$

[Approach 3] Substitute $A=5-B$ into $A^2+B^2$ to get the convex parabola $2B^2 - 10B + 25$ for the objective function which is minimized on it's axis of symmetry, $B= \frac{-(-10)}{2(2)} = \frac{5}{2}$ $\Rightarrow A=\frac{5}{2}$ . Then $9 = A^2 + B^2 +2 A B \cos \theta = 2B^2(1+ \cos \theta ) \Rightarrow \boxed{ \theta = \cos^{-1}( \frac{18}{25} -1) }$ .