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can anyone walk me through how to work this out?
If f(k) = 2k^0.5, s = 0.1, n = 0.1, and d = 0.05, what is the value of f(k) at equilibrium?
If f(k) = 2k^0.5, s = 0.1, n = 0.1, and d = 0.05, what is the value of f(k) at equilibrium?
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(Original post by schlut)
can anyone walk me through how to work this out?
If f(k) = 2k^0.5, s = 0.1, n = 0.1, and d = 0.05, what is the value of f(k) at equilibrium?
can anyone walk me through how to work this out?
If f(k) = 2k^0.5, s = 0.1, n = 0.1, and d = 0.05, what is the value of f(k) at equilibrium?
Hence sf(k)=(n+d)k
0.2k^0.5 = 0.15k
solving for k yields; k= 0 or 16/9, plug into f(k)
Hope this helps
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