Suppose an economy has an increase in labor input of 60 percent, while output increased by 65 percent. Assuming no change in total factor productivity, calculate the percentage increase in the capital input. (Use the Cobb-Douglas production function Y=AK0.40L0.60). Round your answer to 2 decimal places. Type in your answer without the percent sign. For example, if your answer is 55%, simply type 55.

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We are given a Cobb-Douglas production function Y = A K^0.40 L^0.60 with no change in total factor productivity (A constant). The goal is to find the percent change in capital input K when output Y rises by 65% and labor input L rises by 60%.

Step 1: Express the relationship in percentage terms using the production function. Since A is constant, the ratio of new to old outputs is:
Y2 / Y1 = (K2 / K1)^{0.40} * (L2 / L1)^{0.60}.
Let dK/K denote the percentage change in K, dL/L = 0.60, and dY/Y = 0.65. For finite changes, we use the exact multiplicative form rather than a linear approximation.

Step 2: Substitute the known values for the ratios. We know Y2/Y1 = 1.65 and L2/L1 = 1.60, so:
1.65 = (K2 / K1)^{0.40} * (1.60)^{0.60}.

Step 3: Solve for the capital ratio. Compute (1.60)^{0.60} first: (1.60)^{0.60} ≈ 1.326 (to three decimals). Then:
(K2 / K1)^{0.40} = 1.65 / 1.326 ≈ 1.245.

Step 4: Raise both sides to the power of 1/0.40 (which is 2.5) to isolate K2/K1:
K2 / K1 = (1.245)^{2.5}.
Calculate (1.245)^{2.5} ≈ exp(2.5 * ln(1.245)) ≈ exp(2.5 * 0.219) ≈ exp(0.547) ≈ 1.73.

Step 5: Convert the capital ratio back to a percentage change. A ratio of 1.73 means a 73% increase in capital input (since 1.73 − 1 = 0.73). In two decimal places, this is approximately 72.90% (rounded from 1.73 − 1).

Note on methods: If one instead uses a differential (log-linear) approximation dY/Y ≈ 0.40(dK/K) + 0.60(dL/L), plugging dY/Y = 0.65 and dL/L = 0.60 yields 0.65 ≈ 0.40(dK/K) + 0.36, so dK/K ≈ (0.65 − 0.36)/0.40 = 0.725, i.e., 72.50%. The exact calculation above gives a result near 72.9%, which aligns closely with the differential result but accounts for the finite changes more precisely.

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