I know that q= 120L^{1 1/2} K - (L^{3}/√K) has homogenous degree of 2 1/2 but I'm unsure how that has been worked out. I think its the square root thats throwing me, could someone show me how its 2 1/2?
Thanks in advance.
I know that q= 120L^{1 1/2} K - (L^{3}/√K) has homogenous degree of 2 1/2 but I'm unsure how that has been worked out. I think its the square root thats throwing me, could someone show me how its 2 1/2?
Thanks in advance.
In this case, homogeneous means that the sum of the exponents of L and K for the first term is the same as the sum of the exponents of L and K for the second term.
For the first term, L has exponent $\displaystyle 1\,\frac{1}{2}$ and K has exponent 1, so the sum is $\displaystyle 2\,\frac{1}{2}$. For the second term, L has exponent 3 and K has exponent $\displaystyle -\frac{1}{2}$ (since $\displaystyle K^{-\frac{1}{2}}=\frac{1}{\sqrt{K}}$), so the sum is again $\displaystyle 2\,\frac{1}{2}$.
- Hollywood