A Cobb-Douglas production function is given by:
$Q(x,y) = x^{1/2}y^{1/4}$
where x and y are the input variables. The price of each product is p, and the cost of production is $C(x, y) = ax + by$.
I need to write down the profit, Π and find the stationary point for the profit Π and show that the critical values for x and y are given by
$x_* = \frac{p^4}{32a^3b}$ and $y_* = \frac{p^4}{64a^2b^2}$
So I started this question by forming an equation for the profit:
$Π = R - C = pQ(x,y) - (ax + by)$
Then the First Order Conditions are as follows:
$Π_x = \frac{1}{2}px^{-1/2}y^{1/4} - a$,
$Π_y = \frac{1}{4}px^{1/2}y^{-3/4} - b$
Anyone know where to take this from here? Cheers.
As Henry has already commented set the derivatives equal to zero
$$\frac{1}{2}px^{-1/2}y^{1/4} - a=0\Rightarrow \frac{1}{2}px^{-1/2}y^{1/4} =a \quad (1)$$
$$\frac{1}{4}px^{1/2}y^{-3/4} - b=0\Rightarrow \frac{1}{4}px^{1/2}y^{-3/4} = b \quad (2)$$
Divide the first equation by the second equation. The laws of exponents are helpful here.
$$\frac{\frac{1}{2}px^{-1/2}y^{1/4}}{\frac{1}{4}px^{1/2}y^{-3/4} }=\frac{a}{b}$$
$$2\cdot x^{-1/2-1/2}y^{1/4-(-3/4)}=\frac{a}{b}$$
$$2\cdot x^{-1}y^{1}=\frac{a}{b}\Rightarrow 2y=\frac{ax}{b}\Rightarrow y=\frac{ax}{2b} \quad (3)$$
Next we can raise (1) to the fourth power to get an $y^1$:
$$\frac{1}{16}p^4x^{-2}y^{1} =a^4$$
Inserting the expression for $y$
$$\frac{1}{16}p^4x^{-2}\cdot \frac{ax}{2b} =a^4$$
Next you solve the equation for $x$ to obtain $x^*$. Before you start use that $x^{-2}\cdot x=\frac1x$
Finally use $(3)$ and the expression for $x^*$ to calculate $y^*$.