I have the following equation $$Q_1(K, L) = AK^α L^β, A, α, β > 0$$ I'm asked to derive the isoquant curve which yields the following diagram: graph

Could anyone please explain to me how the form of the graph was found?

I have 5 unknowns in this equation and I just can figure out how would it be possible to plot the graphs.

Thank you very much in advance for your help!


Each curve shows the rate at which K can be substituted for L, or vice versa, while keeping output constant. The Marginal Rate of Technical Substitution (MRTS) equals the absolute value of the slope. The MRTS tells us how much one input a firm can sacrifice while still mainting a certain output level. Then the MRTS is equal to the following ratio: $$\frac{\text{Marginal Productivity of Capital } (MP_{K})}{\text{Marginal Productivity of Labour } (MP_{L})}$$

Then the substitution of Labour (L) for Capital (K) is given by: $$ MRTS_{K,L} = -dL/dK = MP_{K}/MP_{L}$$

Now, note that in your equation $\alpha$, $\beta$ and $A$ are exogenous parameters (i.e. given to you). What your equation will then defined is, given these parameters, how your inputs (K and L) result in some amount of output given your production function.

You can find the isoquant curve that yields that diagram by varying the fixed level of output and by "playing with the fixed parameters".

  • $\begingroup$ Thank you very much for your answer! I got it know though how would this apply for a equation similar to $Q_3(K,L) = Amin(B_1K+B_2L,C_1K+C_2L)$ $\endgroup$ – Fozoro Apr 7 '19 at 19:14
  • 1
    $\begingroup$ Well in this case you simply define the function based on the parameters $B_{1}, B_{2}, C_{1}, C_{2}$ (i.e. set $B_{1}K + B_{2}L < C_{1}K + C_{2}L$ and the opposite case to get the conditions in which your function is equal to $B_{1}K + B_{2}L $ or $ C_{1}K + C_{2}L$). Then you differentiate either function based on whether the condition $\frac{K}{L} < \frac{C_{2} - B_{2}}{B_{1}-C_{1}} $ is met or not. (See math.stackexchange.com/questions/2373324/… for further clarification on the derivative of a min function) $\endgroup$ – Ali Apr 7 '19 at 19:25
  • $\begingroup$ Great! thank you very much $\endgroup$ – Fozoro Apr 7 '19 at 19:25

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