Why Input Prices and Marginal Products Must Be Compared
In the long run, a firm can choose how much labor and capital to use. For a specified output target, many combinations of the two inputs may be technically feasible, but their costs can differ. The cost-minimization problem is to produce at least the target output using the least expenditure. The key condition compares the productivity of each input with its price: at an interior optimum, the marginal rate of technical substitution equals the ratio of input prices.
This condition is more than a geometric tangency rule. It follows from asking whether a small change in the input mix can preserve output while lowering expenditure. The derivation also clarifies which way a firm should adjust when the condition is not satisfied.
Let \(w>0\) be the price of one unit of labor and \(r>0\) the rental price of one unit of capital. Total expenditure is \(C=wL+rK\). The firm seeks to minimize this expenditure subject to \(F(L,K)\geq \bar q\), where \(\bar q\) is its required output. When inputs are productive and prices are positive, a cost-minimizing choice will ordinarily produce exactly \(\bar q\): producing more than the target with the same inputs is unnecessary for this constrained problem.
Deriving the Interior Cost-Minimizing Rule
First consider an interior choice, meaning \(L>0\) and \(K>0\), with differentiable production and positive marginal products. At such a choice, small changes in both inputs are available. The Lagrangian method gives the condition directly.
Minimize \(wL+rK\) subject to \(F(L,K)=\bar q\). Use the Lagrangian \(\mathcal{L}=wL+rK+\lambda[\bar q-F(L,K)]\).
At an interior stationary choice, differentiation with respect to each input gives \(w=\lambda MP_L\) and \(r=\lambda MP_K\).
Since \(r>0\), \(MP_K>0\), and the multiplier is positive under these assumptions, dividing eliminates \(\lambda\): \(w/r=MP_L/MP_K\).
The interior condition is \(MRTS_{LK}=w/r\). The output-preserving tradeoff between inputs must match the tradeoff implied by their prices.
The second version says that the marginal product obtained per dollar spent is equal across the two inputs. If one input produced more additional output per dollar than the other, the firm could redirect expenditure toward the more productive input and lower the cost of reaching the target.
Why the Rule Holds Along an Isoquant
The same result can be seen without the Lagrangian. On the isoquant for \(\bar q\), output is fixed, so a small change must satisfy \(dF=0\). The total differential of production is \(dF=MP_L\,dL+MP_K\,dK\). Solving for the required change in capital gives \(dK/dL=-MP_L/MP_K\). The corresponding change in expenditure is therefore:
If \(MRTS_{LK}\neq w/r\), an available small change in the input mix can reduce expenditure: choose the sign of \(dL\) so that \(dC<0\). Thus, at an interior minimum, the coefficient of \(dL\) must be zero, which requires \(MRTS_{LK}=w/r\). This establishes a necessary first-order condition. It is not, by itself, proof that every point satisfying the equality is a minimum. With convex isoquants and the usual regularity conditions, the tangency gives a minimum; otherwise, check the second-order conditions or compare feasible costs.
Worked Examples
Worked Example: Solving for an Interior Input Mix
A hypothetical workshop produces \(F(L,K)=\sqrt{L}+\sqrt{K}\) units of output. It must produce \(\bar q=10\). Labor costs \(w=\$4\) per unit and capital costs \(r=\$1\) per unit. Find the interior cost-minimizing input quantities and verify the rule.
Method. Calculate the marginal products, equate their ratio to the input-price ratio, and use the output constraint to determine the quantities. The function is strictly concave, so its output requirement set is convex; an interior stationary choice is a global cost minimum here.
The marginal products are \(MP_L=1/(2\sqrt{L})\) and \(MP_K=1/(2\sqrt{K})\). Hence \(MRTS_{LK}=\sqrt{K/L}\). The price ratio is \(w/r=4/1=4\), so the rule requires \(\sqrt{K/L}=4\), or \(K=16L\).
Substitute \(K=16L\) into the target-output equation: \(10=\sqrt{L}+\sqrt{16L}=\sqrt{L}+4\sqrt{L}=5\sqrt{L}\). Thus \(\sqrt{L}=2\), \(L=4\), and \(K=16(4)=64\).
Expenditure is \(C=4(4)+1(64)=16+64=\$80\). At these quantities, \(MP_L=1/(2\cdot2)=1/4\) and \(MP_K=1/(2\cdot8)=1/16\). Their ratio is \((1/4)/(1/16)=4=w/r\), and output is \(\sqrt{4}+\sqrt{64}=2+8=10\). Both the target and the first-order condition check.
Common mistake avoided. Equating the marginal products themselves would give \(MP_L=MP_K\), which ignores the different input prices. The relevant comparison is marginal product per dollar, or equivalently \(MP_L/MP_K=w/r\).
Worked Example: Using the Rule and Checking the Minimum
A hypothetical service provider has production function \(F(L,K)=L+\sqrt{K}\) and must produce \(\bar q=10\). The wage is \(w=\$4\), and the rental price of capital is \(r=\$1\). Find its least-cost input combination.
Method. Apply the interior condition, then substitute the output constraint into the cost function to verify that the stationary choice is a minimum. Here \(MP_L=1\), \(MP_K=1/(2\sqrt K)\), and \(MRTS_{LK}=2\sqrt K\).
The price ratio is \(w/r=4\). The equality \(2\sqrt K=4\) gives \(\sqrt K=2\), so \(K=4\). The output constraint then gives \(L+\sqrt4=10\), hence \(L=8\). The choice is interior because both inputs are positive.
To check the minimum, write \(z=\sqrt K\), so \(z\geq0\), \(L=10-z\), and feasibility requires \(0\leq z\leq10\). Expenditure becomes \(C=4(10-z)+z^2=40-4z+z^2=(z-2)^2+36\). This is minimized at \(z=2\), which is feasible, confirming \(K=4\) and \(L=8\). The minimum expenditure is \(C=4(8)+4=\$36\). Output is \(8+2=10\), and the marginal-product ratio is \(1/(1/4)=4=w/r\).
Common mistake avoided. The first-order condition identifies a stationary candidate; it should not be treated as a minimum without checking that the candidate is feasible and minimizes cost. Completing the square verifies the minimum in this example.
Worked Example: Using the Inequality to Choose a Direction
A hypothetical producer has \(F(L,K)=2\sqrt{L}+4\sqrt{K}\). At \(L=16\) and \(K=16\), it produces \(2(4)+4(4)=24\) units. Labor costs \(w=\$2\) per unit and capital costs \(r=\$5\) per unit. Is this input mix cost-minimizing, and which direction along the isoquant lowers cost?
Method. Compare \(MRTS_{LK}\) with \(w/r\), then use the expenditure differential to determine the cost-reducing direction. The marginal products are \(MP_L=1/\sqrt L\) and \(MP_K=2/\sqrt K\). At the stated quantities, \(MP_L=1/4=0.25\) and \(MP_K=2/4=0.50\), so \(MRTS_{LK}=0.25/0.50=0.5\).
The input-price ratio is \(w/r=2/5=0.4\). Since \(0.5>0.4\), the interior equality fails. The isoquant slope is \(dK/dL=-0.5\). For a small increase \(dL>0\), the expenditure change per unit increase in labor is \(dC/dL=w+r(dK/dL)=2+5(-0.5)=-0.5\). Thus, increasing labor and reducing capital along the isoquant lowers cost locally; the original mix cannot be an interior minimum.
As a finite check, raise labor from 16 to 17 and keep output at 24. The output equation gives \(2\sqrt{17}+4\sqrt K=24\), so \(\sqrt K=6-\sqrt{17}/2\) and \(K=161/4-6\sqrt{17}\approx15.5114\). The new cost is \(2(17)+5(15.5114)\approx\$111.5568\), less than the original \(2(16)+5(16)=\$112\). Output remains 24 by construction.
Common mistake avoided. Reversing the ratio or calculating a marginal product incorrectly can reverse the recommended adjustment. Here \(MP_K=2/\sqrt{16}=0.50\), not \(0.25\); using the correct value gives \(MRTS_{LK}=0.50\) and the cost-reducing direction stated above.
Interpreting the Condition and Its Limits
The comparison between \(MRTS_{LK}\) and \(w/r\) gives a practical adjustment rule for an interior point. From the isoquant expenditure differential, if \(MRTS_{LK}>w/r\), then a small increase in labor, with the corresponding reduction in capital, lowers cost. If \(MRTS_{LK}<w/r\), a small decrease in labor, with the corresponding increase in capital, lowers cost. At an interior optimum neither direction can lower cost, so the two ratios are equal.
| Comparison at a feasible interior point | Cost-reducing movement along the isoquant |
|---|---|
| \(MRTS_{LK}>w/r\) | Increase \(L\) and reduce \(K\) |
| \(MRTS_{LK}<w/r\) | Reduce \(L\) and increase \(K\) |
| \(MRTS_{LK}=w/r\) | No first-order cost reduction from a small reallocation |
The rule presumes a smooth isoquant and an interior choice. A firm at a boundary cannot necessarily make a small change in both directions. A kink may not have a single slope, and zero marginal products prevent the ratio formula from being used as written. These cases require analyzing feasible choices directly rather than mechanically imposing the equality.
A further exam trap is to confuse a necessary condition with a complete solution. An equality can identify a candidate, but the candidate must also meet the output requirement and be feasible; the shape of the technology or a direct cost comparison must support that it is a minimum. When the problem provides convex isoquants and an interior solution, the equality is the central input-mix condition. For technologies with different properties, check the conditions stated in the problem.
Why the Rule Matters
The cost-minimizing input rule connects production technology to factor markets. Marginal products describe the technical contribution of each input, while \(w/r\) describes their relative cost. The firm’s least-cost mix balances those two comparisons for its required output. This logic underlies long-run cost calculations and decisions about how a producer allocates expenditure between inputs.
Check Your Understanding
For each problem, identify the relevant input-price ratio and apply the output constraint where needed.
- A producer has \(F(L,K)=\sqrt L+2\sqrt K\). At \(L=4,K=1\), labor costs \(w=3\) and capital costs \(r=1\). Calculate both marginal products and the MRTS, compare it with the price ratio, and state the cost-reducing direction. (Answer: \(MP_L=0.25\), \(MP_K=1\), \(MRTS_{LK}=0.25<3=w/r\); reduce labor and increase capital.)
- A firm produces \(F(L,K)=2\sqrt L+\sqrt K\), with target output 9, \(w=2\), and \(r=1\). Find the interior cost-minimizing quantities and expenditure. (Answer: \(L=9\), \(K=9\), minimum expenditure \(=\$27\).)
- A producer has \(F(L,K)=L+\sqrt K\), target output 14, wage \(w=2\), and rental price \(r=1\). Use the rule and verify the minimum by writing cost as a function of \(\sqrt K\). (Answer: \(K=1\), \(L=13\), minimum expenditure \(=\$27\), since \(C=(\sqrt K-1)^2+27\).)
- A firm uses \(F(L,K)=\sqrt L+\sqrt K\) to produce 5 units. Labor costs \(w=1\) and capital costs \(r=4\). Solve for the interior cost-minimizing quantities and check both output and the marginal-product ratio. (Answer: \(L=16\), \(K=1\), minimum expenditure \(=\$20\), with \(MRTS_{LK}=1/4=w/r\).)