Science:Math Exam Resources/Courses/MATH105/April 2011/Question 08 (a)/Solution 1

The marginal productivity of labour is the rate of change of productivity with respect to the labour investment, so in this case, it corresponds to the partial derivative

$\displaystyle {\begin{aligned}{\frac {\partial f}{\partial x}}(x,y)&={\frac {\partial }{\partial x}}(160x^{3/4}y^{1/4})\\&=160{\frac {3}{4}}x^{-1/4}y^{1/4}\\&=120x^{-1/4}y^{1/4}\end{aligned}}$

And the marginal productivity of capital is the rate of change of productivity with respect to the capital investment, so in this case, it corresponds to the partial derivative

$\displaystyle {\begin{aligned}{\frac {\partial f}{\partial y}}(x,y)&={\frac {\partial }{\partial y}}(160x^{3/4}y^{1/4})\\&=160x^{3/4}{\frac {1}{4}}y^{-3/4}\\&=40x^{3/4}y^{-3/4}\end{aligned}}$

The question asks us to compute those marginal productivities for 81 units of labour and 16 units of capital, that is to compute

$\displaystyle {\frac {\partial f}{\partial x}}(81,16)\quad {\text{and}}\quad {\frac {\partial f}{\partial y}}(81,16)$

which yields

$\displaystyle {\begin{aligned}{\frac {\partial f}{\partial x}}(81,16)&=120\cdot 81^{-1/4}16^{1/4}\\&=120\cdot {\frac {1}{3}}\cdot 2\\&=80\end{aligned}}$

and

$\displaystyle {\begin{aligned}{\frac {\partial f}{\partial y}}(81,16)&=40\cdot 81^{3/4}16^{-3/4}\\&=40\cdot 27\cdot {\frac {1}{8}}\\&=135\end{aligned}}$

So the marginal productivity of labour is 80 (dollars of productivity per dollar of labour investment) and the marginal productivity of capital is 135 (dollars of productivity per dollar of capital investment).