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I was wondering can anyone show step by step how this is done

Determine whether constant, decreasing, or increasing returns to scale

1) F(K,L) = K^2/L

2) F(K,L) = K+L

3) F(K,L)= square root of K+ square root of L

4) F(K,L)= K^2+L^2

Fg will be given to anyone who gives a clear step by step.

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You just have to compare aF(K,L) with F(aK,aL) where a is a constant greater than 1.

If aF(K,L) = F(aK,aL) it is constant EDIT: in the constant case a >=0

If aF(K,L) > F(aK,aL) it is decreasing

If aF(K,L) < F(aK,aL) it is increasing

1) F(K,L) = K^2/L

F(aK,aL) = a^2 K^2/aL = aK^2/L

aF(K,L) = aK^2/L

Constant

2) F(K,L) = K+L

F(aK,aL)= a + K + a + L = 2a + K + L

aF(K,L) = a(K+L)

Pretty sure this is decreasing

3) F(K,L)= square root of K+ square root of L

F(aK,aL)= root aK + root aL = roota (rootk + rootl)

aF(K,L) = a(rootK+rootL)

Decreasing

4) F(K,L)= K^2+L^2

F(aK,aL)= a^2 K^2 + a^2 L^2 = a^2(K^2+L^2)

aF(K,L) = a(K^2+L^2)

Increasing.

Pretty sure I remembered this correctly. Hope it helps!

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I still dont get it lol.

Why is #2 decreasing?

2) F(K,L) = K+L

F(aK,aL)= a + K + a + L = 2a + K + L

aF(K,L) = a(K+L)

I thought you only put the constant a in front of the K and L. So it'll be like F(K,L) = aK+aL

hmm would it be better if i just substituted a value for K, L, and a? and then I can just add them up and compare the original F(K,L) to the new F(K2,L2)?

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I still dont get it lol.

Why is #2 decreasing?

2) F(K,L) = K+L

F(aK,aL)= a + K + a + L = 2a + K + L

aF(K,L) = a(K+L)

I thought you only put the constant a in front of the K and L. So it'll be like F(K,L) = aK+aL

hmm would it be better if i just substituted a value for K, L, and a? and then I can just add them up and compare the original F(K,L) to the new F(K2,L2)?

I don't understand what you're saying, but I believe it is decreasing because aF(K,L) is multiplicative as opposed to being additive and a > 1 so that function will eventually dominate and be the upper function.

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• 1 year later...

this looks like a good place for me to spam

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This is actually a terrible place to spam.