Perfect Substitutes and Corner Input Choices

Why Input Choice Can Have a Corner Solution

A firm choosing inputs normally asks how to produce a target quantity of output at the lowest possible cost. When labor and capital are perfect substitutes, the production technology allows one input to replace the other at a constant rate. If one input produces more output per dollar, the least-cost plan uses that input alone. If the inputs produce the same output per dollar, more than one mix can attain the minimum cost.

This is a useful contrast with cost-minimization problems in which an optimum is found by matching the slope of an isoquant to the slope of an isocost. With perfect substitutes, the isoquants are straight lines, so their slope does not change from one input mix to another. The cost-minimizing solution can therefore be at a boundary—using only one input—rather than at an interior tangency.

Definition: A production function with perfect substitutes has the linear form $q=aL+bK$, where $q$ is output, $L$ is labor, $K$ is capital, and $a,b>0$ are constant amounts of output produced per unit of each input. Inputs are divisible and nonnegative. If labor costs $w>0$ per unit and capital costs $r>0$ per unit, total input cost is $C=wL+rK$.

The model treats each unit of labor as adding $a$ units of output and each unit of capital as adding $b$ units, regardless of how much of either input is already used. The firm chooses $L$ and $K$ to meet a specified target $q$ at minimum cost. There are no fixed setup costs, quantity constraints, or other production requirements in this model. Those assumptions matter: a fixed cost for using an input or a limit on how much can be purchased could change the best choice.

Compare Output per Dollar

The relevant comparison is not which input has the higher productivity by itself, or which has the lower price by itself. Instead, calculate the output obtainable from one dollar spent on each input. Labor produces $a$ units per unit at price $w$, so its output per dollar is $a/w$. Capital’s output per dollar is $b/r$. Equivalently, compare the cost of obtaining one unit of output from each input: $w/a$ for labor and $r/b$ for capital.

\begin{aligned} \text{Labor output per dollar} &= \frac{a}{w}, &\qquad \text{labor cost per output unit} &= \frac{w}{a},\\ \text{Capital output per dollar} &= \frac{b}{r}, &\qquad \text{capital cost per output unit} &= \frac{r}{b}. \end{aligned}

If $a/w>b/r$, labor produces more output per dollar, so the least-cost plan uses labor alone. If $a/w<b/r$, capital produces more output per dollar, so the least-cost plan uses capital alone. If the ratios are equal, neither input has a cost advantage per unit of output. Any nonnegative mix that produces exactly the target output has the same minimum cost.

Theorem: For a positive target $q$, a linear production function $q=aL+bK$, and positive input prices $w,r$, the minimum cost is $C(q)=q\min\left(\frac{w}{a},\frac{r}{b}\right)$. If $a/w>b/r$, a cost-minimizing plan is $L=q/a,\ K=0$. If $a/w<b/r$, it is $L=0,\ K=q/b$. If $a/w=b/r$, every nonnegative input mix satisfying $aL+bK=q$ is cost-minimizing.

Why the Comparison Gives the Minimum

The result follows by expressing each input’s contribution to output in terms of its cost. Labor quantity $L$ contributes $aL$ units of output and costs $wL$, or $w/a$ dollars per output unit. Capital quantity $K$ contributes $bK$ units and costs $rK$, or $r/b$ dollars per output unit. Since total output is the sum of the two contributions, total cost can be written as the sum of the cost of producing each contribution.

C=wL+rK =\frac{w}{a}(aL)+\frac{r}{b}(bK).

The two output contributions, $aL$ and $bK$, are nonnegative and add to $q$. Therefore, each dollar cost per output unit is being applied to part of the target. The total cost cannot be less than the lower of the two unit costs multiplied by the whole target:

C =\frac{w}{a}(aL)+\frac{r}{b}(bK) \geq \min\left(\frac{w}{a},\frac{r}{b}\right)(aL+bK) =q\min\left(\frac{w}{a},\frac{r}{b}\right).

This lower bound is attainable: produce all $q$ units using whichever input has the lower cost per output unit. If the two unit costs are equal, every output contribution has that same cost per unit, so any mix producing exactly $q$ reaches the bound. Positive input prices also explain why an optimal plan does not produce more than the target: extra input use would add cost without helping meet the requirement.

Calculate output per dollar.
Compute $a/w$ for labor and $b/r$ for capital, keeping each price attached to the correct input.

Compare the two ratios.
The larger output-per-dollar ratio identifies the input with the lower cost per output unit. Equal ratios indicate a tie.

Meet the target and calculate cost.
Use only the more cost-effective input when the ratios differ. When they tie, describe the full set of mixes that produce the target.

Worked Cost-Minimization Problems

Worked Example: Labor Produces More Output per Dollar

A workshop has production function $q=4L+3K$. Labor costs $w=\$12$ per unit and capital costs $r=\$15$ per unit. Find the least-cost input plan for a target of 240 output units, calculate its cost, and check the result.

Method: compare output per dollar. Labor produces $a=4$ units per labor unit and capital produces $b=3$ units per capital unit. Their output per dollar is $a/w=4/12=1/3$ for labor and $b/r=3/15=1/5$ for capital. Since $1/3>1/5$, labor is the lower-cost input per output unit.

The labor-only quantity needed is $L=q/a=240/4=60$, with $K=0$. Its cost is $C=12(60)+15(0)=\$720$. The unit-cost comparison gives the same result: labor costs $w/a=12/4=\$3$ per output unit, so $240(3)=\$720$.

Check. Production is $4(60)+3(0)=240$, exactly the target. As a comparison, capital alone would require $K=240/3=80$ and cost $15(80)=\$1{,}200$, which is higher than \$720.

Mistake avoided: Choosing the input with the larger productivity coefficient is not a complete method. Here the price-adjusted comparison confirms that labor is best; productivity must be evaluated relative to price.

Worked Example: The Cheaper Input Is Not the Least-Cost Input

A packaging firm’s production function is $q=2L+5K$. Labor costs $w=\$10$ per unit and capital costs $r=\$20$ per unit. Find the cost-minimizing plan for 300 output units and compare it with using labor alone.

Method: compare output per dollar rather than nominal input prices. Labor produces $2/10=0.2$ output units per dollar. Capital produces $5/20=0.25$ output units per dollar. Although labor has the lower price per input unit, $0.25>0.2$, so capital produces more output per dollar.

Using capital alone, the required quantity is $K=q/b=300/5=60$, with $L=0$. The cost is $C=10(0)+20(60)=\$1{,}200$. Capital’s unit cost of output is $r/b=20/5=\$4$, and $300(4)=\$1{,}200$.

Using labor alone would require $L=300/2=150$. Its cost would be $10(150)=\$1{,}500$. Production under the chosen plan checks as $2(0)+5(60)=300$; the plan meets the target and costs \$300 less than the labor-only alternative.

Mistake avoided: Selecting labor simply because its input price is lower would confuse price per input unit with cost per output unit. Capital is more expensive per unit but also more productive, and the output-per-dollar comparison accounts for both facts.

Worked Example: Equal Output per Dollar and Multiple Optimal Mixes

A small producer has production function $q=3L+2K$. Labor costs $w=\$15$ per unit, and capital costs $r=\$10$ per unit. Find the minimum cost for 360 output units and describe the full set of cost-minimizing mixes. Also verify one mixed-input plan.

Method: test for equality of output per dollar. Labor produces $3/15=0.2$ output units per dollar, and capital produces $2/10=0.2$ output units per dollar. Both have a cost of $\$5$ per output unit: $15/3=5$ for labor and $10/2=5$ for capital. Thus the minimum cost is $360(5)=\$1{,}800$.

Every nonnegative mix satisfying $3L+2K=360$ is optimal. Solving this constraint for capital gives $K=(360-3L)/2=180-1.5L$. For $K\geq0$ and $L\geq0$, the range is $0\leq L\leq120$, with $K=180-1.5L$.

For example, take $L=80$. Then $K=180-1.5(80)=180-120=60$. Production is $3(80)+2(60)=240+120=360$, and cost is $15(80)+10(60)=\$1{,}200+\$600=\$1{,}800$. At the endpoints, labor alone requires $L=120$ and costs $15(120)=\$1{,}800$; capital alone requires $K=180$ and costs $10(180)=\$1{,}800$.

Mistake avoided: Reporting only the two endpoint plans would miss the equally optimal mixed plans. Equality of output per dollar means the whole line of nonnegative mixes meeting the target has minimum cost.

Common Mistakes and Exam Traps

Comparing productivity without price. A larger $a$ or $b$ alone does not identify the least-cost input. Compare $a/w$ with $b/r$, or equivalently compare $w/a$ with $r/b$.
Reversing the ratios. Larger output per dollar is better, but smaller cost per output unit is better. These are reciprocal comparisons, so the preferred input is identified by opposite inequality directions.
Assuming every optimum is a tangency. A linear production function has a constant trade-off between inputs. If their cost per output unit differs, the optimal plan is a corner using only the more cost-effective input.
Calling a tie a unique solution. When $a/w=b/r$, the minimum cost is unique for a given target, but the input mix generally is not. Describe the feasible line of optimal mixes, not only one example.
Using excess inputs without a reason. With positive prices and a requirement to produce at least a target, an optimal plan meets the target exactly. Producing extra output adds cost and is unnecessary.

Key takeaway: For perfect substitutes, the least-cost choice is determined by output per dollar. A strict advantage leads to an all-labor or all-capital solution; equal output per dollar allows every nonnegative mix on the target-output line to be optimal.

Check Your Understanding

For each problem, identify the cost-minimizing input choice or set of choices and check the stated target against the production function.

A production function is $q=5L+2K$, with input prices $w=\$25$ and $r=\$20$. Find the least-cost plan and cost for $q=100$. (Answer: $L=20,\ K=0$; minimum cost $\$500$.)
A firm produces $q=4L+3K$, with $w=\$16$ and $r=\$9$. Find the least-cost plan for $q=180$ and the cost of using labor alone for the same target. (Answer: Use $K=60,\ L=0$, at cost $\$540$; labor alone costs $\$720$.)
A production function is $q=2L+6K$, with $w=\$12$ and $r=\$36$. For a target of 120, determine whether the optimum is unique and give one mixed-input optimum. (Answer: It is not unique; every nonnegative mix with $2L+6K=120$ costs $\$720$, including $L=30,\ K=10$.)
A firm has $q=7L+4K$, with $w=\$21$ and $r=\$16$. Determine the least-cost plan and minimum cost for $q=280$. (Answer: Use $L=40,\ K=0$; minimum cost $\$840$.)
A production function is $q=0.5L+2K$, with $w=\$5$ and $r=\$20$. For a target of 90, describe the optimal mixes and verify the mix $L=100,\ K=20$. (Answer: Both inputs produce $0.1$ output units per dollar; all nonnegative mixes with $0.5L+2K=90$ cost $\$900$, including $L=100,\ K=20$.)

What You'll Learn