Boundary Solutions with Nonstandard Technologies

Why Boundary Comparisons Matter

For a smooth technology with a well-behaved, convex isoquant, an interior tangency between an isoquant and an isocost line can help identify the least-cost input combination. But that method does not settle every production problem. A technology may offer separate production modes, have a nonconvex isoquant, change output abruptly, or restrict feasible inputs. In such cases, a tangency can be absent, can occur at a non-minimum, or can miss a cheaper boundary bundle.

The reliable question is always: among all bundles that produce at least the target output, which has the lowest cost? This requires checking feasibility and comparing costs, not merely solving a derivative condition.

Definition: Let $K\geq 0$ and $L\geq 0$ denote two inputs, let $r>0$ and $w>0$ be their respective prices, and let $f(K,L)$ be output. For a target output $\bar q$, the feasible set is the set of all input bundles satisfying $f(K,L)\geq \bar q$. The cost of a bundle is $C(K,L)=rK+wL$. A cost-minimizing bundle is a feasible bundle with the lowest cost among all feasible bundles. A boundary solution lies on the edge of the feasible input set or technology; a corner solution is a boundary solution such as $K=0$ or $L=0$.

This setup assumes input prices are positive and inputs cannot be negative. Output at least equal to the target is sufficient; producing more than the target is permitted. A minimum exists only if some feasible bundle actually attains the lowest cost. If feasible costs get arbitrarily close to a value without reaching it, that value is an infimum, not a minimum.

Compare Feasible Sets, Not Just Tangencies

For any proposed bundle, first verify that it meets the output target, then calculate its cost. A bundle that looks efficient but fails the output requirement is not a candidate. Likewise, a low-cost bundle in one production mode does not establish the overall minimum until other feasible modes have been considered.

A useful organizing principle applies when the feasible technology consists of separate branches. Find the lowest cost within each branch, then compare those branch minima. The global minimum is the smallest of them, provided each branch minimum is attained. This follows directly because every feasible bundle belongs to at least one branch: its cost cannot be lower than the minimum cost for that branch.

Worked Example: Choosing Between Two Production Branches

A workshop can use either of two production modes. Its output is $f(K,L)=\max\{2K+L,\ K+3L\}$. The target is $\bar q=12$, capital costs $r=\$3$ per unit, and labor costs $w=\$2$ per unit. Find the global least-cost bundle.

Method. Treat each production mode as a separate feasible branch, find that branch’s least cost, and compare the results. The feasible condition is $2K+L\geq 12$ or $K+3L\geq 12$; it is not necessary for both inequalities to hold.

Branch 1: $2K+L\geq 12$. Since $K,L\geq 0$, $3K+2L\geq 1.5(2K+L)\geq 1.5(12)=18$. Thus branch 1 costs at least $\$18$. The bundle $K=6,L=0$ produces $2(6)+0=12$ and costs $3(6)+2(0)=\$18$, so this branch minimum is attained.

Branch 2: $K+3L\geq 12$. Here, $3K+2L\geq \frac{2}{3}(K+3L)\geq \frac{2}{3}(12)=8$. The bundle $K=0,L=4$ produces $0+3(4)=12$ and costs $3(0)+2(4)=\$8$. This branch minimum is also attained.

The global minimum is the smaller branch minimum, $\$8$, at $(K,L)=(0,4)$. Check the original technology: $f(0,4)=\max\{4,12\}=12$, so the bundle is feasible. The method avoids assuming that a smooth interior tangency must exist or that each mode must use positive quantities of both inputs.

A Tangency Can Be a Maximum

A first-order condition identifies a stationary point; by itself, it does not establish that the point is a minimum. When an isoquant is nonconvex, an interior stationary point can instead be a local maximum along the isoquant. Boundary candidates must still be compared.

Worked Example: Testing a Stationary Point Against the Corners

Suppose output is $f(K,L)=K^2+L^2$, with $K,L\geq 0$. The target is $\bar q=25$, and input prices are $r=\$3$ and $w=\$2$. Find the least-cost bundle and classify the interior stationary point on the target boundary.

Step 1: Restrict attention to the target boundary. Feasibility requires $K^2+L^2\geq 25$. Because both input prices are positive, reducing a bundle’s inputs while retaining feasibility cannot raise its cost. Thus a cost minimum must be on $K^2+L^2=25$; any bundle strictly outside that boundary can be scaled toward the origin until it reaches the boundary at lower cost.

Step 2: Write cost along the boundary. For $0\leq K\leq 5$, $L=\sqrt{25-K^2}$, so $C(K)=3K+2\sqrt{25-K^2}$. The endpoints are $K=0,L=5$, with cost $3(0)+2(5)=\$10$, and $K=5,L=0$, with cost $3(5)+2(0)=\$15$.

Step 3: Check the interior stationary point. Differentiating along the boundary gives $C'(K)=3-\frac{2K}{\sqrt{25-K^2}}$. Setting this equal to zero gives $3=2K/L$, so $K/L=3/2$. Combining this ratio with $K^2+L^2=25$ yields $K=15/\sqrt{13}$ and $L=10/\sqrt{13}$. Its cost is $3(15/\sqrt{13})+2(10/\sqrt{13})=65/\sqrt{13}$, approximately $\$18.03$.

Step 4: Classify and compare. Along the arc, $C''(K)=-50/(25-K^2)^{3/2}<0$, so the stationary point is a local maximum, not a minimum. The least of the two boundary endpoint costs is $\$10$, attained at $(K,L)=(0,5)$. Check: $f(0,5)=25$ and $C(0,5)=\$10$. The trap is treating the stationary condition as proof of a minimum; the curvature and endpoint comparisons show why that conclusion would be wrong.

This example’s production function has a nonconvex target set: points outside a quarter-circle meet the target, and the arc itself is not a standard convex isoquant. The calculation illustrates why the direction of curvature and the costs at all relevant boundaries matter. A first-order condition is a candidate-generation tool, not a global optimality test.

Discontinuous Technologies and Attainment

With a discontinuity, feasibility can change abruptly as an input crosses a threshold. The least-cost feasible bundle may sit exactly at that threshold. But whether the threshold itself is feasible matters: a condition of $L\geq a$ includes the boundary $a$, whereas $L>a$ excludes it. That small difference can determine whether a cost minimum exists.

Worked Example: An Output Jump with an Attained Minimum

A hypothetical production process uses labor only. Output is $f(L)=2$ for $0\leq L<3$, and $f(L)=9$ for $L\geq 3$. The output target is $\bar q=7$, and labor costs $\$5$ per unit. Find the least-cost feasible input and explain the role of the discontinuity.

Method. Translate the output target into a feasible-input condition before minimizing cost. For every $L<3$, output is $2<7$, so those labor quantities are infeasible. At $L=3$, output jumps to $9$, which meets the target. Therefore the feasible quantities are exactly $L\geq 3$.

Cost is $C(L)=5L$. Since $5>0$, cost increases with $L$ on the feasible set. The smallest feasible quantity, $L=3$, is therefore the cost minimum, and its cost is $5(3)=\$15$. Check: $f(3)=9\geq 7$, while every $L<3$ fails the target. The minimum is attained at the boundary $L=3$, even though the technology is discontinuous there.

A related case has no attained minimum. Suppose output is zero for $L\leq 4$ and ten for $L>4$, with target output eight and labor price $\$6$. Feasible quantities must satisfy $L>4$, so every feasible bundle costs more than $6(4)=\$24$. For any $\epsilon>0$, $L=4+\epsilon$ is feasible and costs $\$24+6\epsilon$. These costs can be made arbitrarily close to $\$24$, but $L=4$ itself is infeasible. Thus the infimum is $\$24$, and there is no cost-minimizing bundle. Reporting $\$24$ as an attained minimum would be incorrect.

A Practical Test for Boundary Minima

When the usual smooth tangency approach is unavailable or inconclusive, organize the calculation as a comparison of feasible candidates. This procedure applies to corners, separate technology branches, and thresholds.

State the target and prices.
Write the required output and each input’s positive price, and state any restrictions on feasible inputs.

Translate output into feasibility.
Identify which bundles meet $f(K,L)\geq\bar q$. For a discontinuous or branch-based technology, keep track of whether a boundary is included and whether the condition is “and” or “or.”

List plausible candidates.
Check feasible corners, endpoints, branch minima, thresholds, and any smooth stationary points. A stationary point is only a candidate until classified and compared.

Compute and compare costs.
Evaluate $rK+wL$ at each feasible candidate, then verify that the lowest cost is attainable. If costs only approach a boundary value, report an infimum rather than a minimum.

Positive input prices help rule out unnecessary inputs when a feasible reduction is possible, but they do not make every feasible set convex or closed. A boundary comparison is only complete when the listed candidates cover all ways to reach the lowest cost. For example, in a branch problem, a separate lower bound for each branch establishes that no unexamined point within that branch can be cheaper.

Common Exam Traps

Assuming every optimum is an interior tangency. An input can optimally be zero, or the production set can have a kink, a jump, or no smooth boundary at all.
Calling every stationary point a minimum. Check the second derivative or otherwise classify the point, then compare it with feasible boundaries.
Confusing “and” with “or.” A technology that produces the maximum of two production modes meets the target if either mode does; requiring both can discard feasible bundles.
Ignoring whether a boundary is included. The conditions $L\geq a$ and $L>a$ can produce different answers: the first may attain a minimum at $a$, while the second may have only an infimum.
Comparing cost without checking output. A cheap bundle is irrelevant if it does not meet the target. Verify the original production condition at the proposed solution.

Key takeaway: For nonstandard technologies, cost minimization is a global feasible-set comparison. Find the relevant branches and boundaries, verify each candidate meets the output target, calculate its cost, and check whether the lowest value is actually attained.

Check Your Understanding

For each problem, identify the feasible boundary or branches before comparing costs.

A technology has output $f(K,L)=\max\{2K+L,\ K+2L\}$, target $\bar q=10$, and input prices $r=\$3$, $w=\$2$. Find the least-cost bundle and verify its output. (Answer: $K=0,L=5$, cost $\$10$, output $10$.)
Output is $K^2+L^2$, the target is $16$, and input prices are $r=\$5$, $w=\$3$. Compare the two axis bundles and state the least-cost bundle. (Answer: $K=0,L=4$, cost $\$12$, output $16$.)
A labor-only process produces $1$ unit when $L<2$ and $6$ units when $L\geq 2$. The target is $5$, and labor costs $\$8$ per unit. Identify the minimum and check that it is attained. (Answer: $L=2$, cost $\$16$, attained because output at $L=2$ is $6$.)
A labor-only process produces zero output when $L\leq 5$ and $12$ units when $L>5$. The target is $10$, and labor costs $\$4$ per unit. State whether there is a minimum and give the infimum if appropriate. (Answer: No minimum; the cost infimum is $\$20$, not attained.)
For $f(K,L)=K^2+L^2$, target $\bar q=25$, and prices $r=\$3,w=\$2$, the interior stationary point on the target arc satisfies $3=2K/L$. Compare its cost with both axis bundles and identify the global minimum. (Answer: The stationary point costs $65/\sqrt{13}\approx \$18.03$, while the axis costs are $\$10$ at $(0,5)$ and $\$15$ at $(5,0)$; the global minimum is $(0,5)$ at $\$10$.)

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