Explanation Given information: Cobb-Douglas function Q = b K α L 1 − α Calculation: Cobb-Douglas function is given as, Q = b K α L 1 − α Differentiate Q with respect to K , keeping L constant, we get ∂ Q ∂ K = ∂ ∂ K ( b K α L 1 − α ) = b L 1 − α ∂ ∂ K ( K α ) = b L 1 − α ( α K α − 1 ) = α b L 1 − α K α − 1 Differentiate Q with respect to L , keeping K constant, we get ∂ Q ∂ L = ∂ ∂ L ( b K α L 1 − α ) = b K α ∂ ∂ L ( L 1 − α ) = b K α [ ( 1 − α ) L 1 − α − 1 ] = b K α ( 1 − α ) L − α = b ( 1 − α ) K α L − α Now the equation is given as, K ∂ Q ∂ K + L ∂ Q ∂ L = Q Substituting the values in the left hand side of the equation, we get K ∂ Q ∂ K + L ∂ Q ∂ L

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Author: DA

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Chapter 8.4, Problem 41P

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The Cobb-Douglas function Q=bKαL1−α satisfies the equation K∂Q∂K+L∂Q∂L=Q.

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Chapter 8.4, Problem 40P

Chapter 8.4, Problem 42P

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The Cobb-Douglas function Q = b K α L 1 − α satisfies the equation K ∂ Q ∂ K + L ∂ Q ∂ L = Q .

Solution Summary: The author explains the Cobb-Douglas function, which satisfies the equation Kpartial Q+L

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The Cobb-Douglas function Q=bKαL1−α satisfies the equation K∂Q∂K+L∂Q∂L=Q.

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Chapter 8 Solutions

The Cobb-Douglas function Q = b K α L 1 − α satisfies the equation K ∂ Q ∂ K + L ∂ Q ∂ L = Q .

Solution Summary: The author explains the Cobb-Douglas function, which satisfies the equation Kpartial Q+L

Videos

To show: The Cobb-Douglas function Q=bKαL1−α satisfies the equation K∂Q∂K+L∂Q∂L=Q.

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Chapter 8 Solutions

To show:

The Cobb-Douglas function Q=bKαL1−α satisfies the equation K∂Q∂K+L∂Q∂L=Q.