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$$F(K,AL) = \left [ \alpha K^{\rho} + (1-\alpha) (AL)^{\rho} \right ]^{\frac{1}{\rho}},$$ So $K$ and $AL$ are the variables here. $lim_{K\rightarrow 0}\dfrac{\partial F}{\partial K}=lim_{K\rightarrow 0}\alpha K^{\rho -1} \left [ \alpha K^{\rho} + (1-\alpha) (AL)^{\rho} \right ]^{\frac{1-\rho}{\rho}}=0\cdot[0+(1-\alpha)(AL)^\rho]^{\frac{1-\rho}{\rho}}$ ...

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Let $x$ and $y$ be the purchase prices of the two bicycles respectively. The profit in the first case then is $p_1=(1.1-1)x+(1.2-1)y=0.1x+0.2y$ $1.1x$ and $1.2y$ are the selling prices. The profit in the second case is $p_2=(1.2-1)x+(1.1-1)y=0.2x+0.1y$ $p_2$ is $RS \ 5$ greater than $p_1$. Therefore $p_2=p_1+5$. It follows $0.2x+0.1y=0.1x+0.2y+5 \quad ... 1 Go back and review and upgrade your Calc 1 and 2. These course are foundational. You are building on a house of cards if you lack comfort here. Get a Schaum's Outline or some programmed workbook. You can't express supply and demand functions without them. Linear Algebra (lots of applications in econometrics) Operations Research (very business oriented ... 1 Let the cost of the shoe be x , y , z in country A, B , C respectively in 1990. In 1995, they became 5x, 9y, 13z due to inflation(that's what the index meant). Now, the difference is 4x, 8y, 12z which is all the same 72$ . From this, x=18 ; y= 9 ; z=6 ; 5x=90; 9y = 81; 13z= 78 ; ans- 90:81:78 = 30:27:26

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If you manipulate the expression you get $$P = A\dfrac{\left[(1+i)^n - 1\right]}{i(1+i)^n}$$ $$(1+i)^n(\frac{Pi}{A}) = (1+i)^n-1$$ $$(1+i)^n \left[1-\frac{Pi}{A}\right] = 1$$ $$(1+i)^n = \dfrac{1}{\left[1-\frac{Pi}{A}\right]}$$ Taing log on both sides We get $$n log(1+i) = log\left(\dfrac{1}{\left[1-\frac{Pi}{A}\right]}\right)$$ n = ...

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