Attempting to solve a practice problem, not sure if I'm heading in the right direction since my solution seems pretty messy. Given the following utility function,

$u(x,y)=min\{x^{1/2},2y\}$, find the Marshallian demands.

My answer:

Since Leontief is perfect complements, must be the case that $x^{1/2}=2y$, substituting this into a budget constraint yields the following:

$p_x \times x + p_y \times y = w$, where w is total income. Taking $x^{1/2}=2y$ and squaring this yields $x=4y^2$. Subbing this into constraint would give:

$p_x \times 4y^2 + p_y \times y = w$, at this point I applied the quadratic formula and got a demand function for y as follows,

$$y = \frac{-p_y \pm \sqrt{p_y^2 + 16 p_x w}}{8p_x}$$

This seems messy to me, I figure I can rule out the minus side of the quadratic seems that would imply y is negative. Even if some confirmation that this is the right approach would be appreciated.


  • $\begingroup$ You can rule out that y is negative because prices for scarce goods are positive. Otherwise they wouldn't be scarce. They would be free. You can also rule out that they are zero because if they would be then there wouldn't be a problem to begin with. $\endgroup$
    – Toby
    Apr 29, 2017 at 7:21
  • $\begingroup$ Have you considered accepting an answer (any)? $\endgroup$
    – luchonacho
    Aug 23, 2017 at 12:25

2 Answers 2


As Amit said, your answer is correct. As Toby said, you can rule out negative output.

There is no much insight to add as an answer really. However, I was also fairly puzzled that such a simple problem gives a complex answer. As it turns out, you get a linear demand (on income) only if the exponents on both components is the same.

Consider the more general case:

$$ U(x,y) = \text{min}\{ax^\alpha,by^\beta\} $$

Optimality requires (* notation for optimal values omitted)

$$ ax^\alpha = by^\beta\ $$

Replacing one of these in the budget constraint leads to:

$$ p_x x + p_y \left(\frac{a}{b}\right)^{\frac{1}{\beta}} x^{\frac{\alpha}{\beta}} = w $$

Here we see that only when $\alpha=\beta$, demand will be linear (on income). In this case:

$$ x = \frac{w}{p_x + p_y\left(\frac{a}{b}\right)^{\frac{1}{\alpha}}} $$

It seems an algebraic solution for the general (unrestricted) case cannot be derived, as the polynomial you get might have irrational powers.


An interesting exploration here is whether there are conditions under which such a demand function reflects a negative relation between price and quantity demanded. Normalizing $p_x =1$ and keeping income constant we have

$$y = \frac{-p_y}{8} + {\sqrt{(p_y/8)^2 + \frac14 w}}$$

$$\implies \frac {\partial y}{\partial p_y} = -\frac {1}{8}+ \frac 12\cdot\left((p_y/8)^2 + \frac14 w\right)^{-1/2}\cdot \frac {2p_y}{8^2}$$

$$= -\frac 18 \cdot \left[ 1- \left((p_y/8)^2 + \frac14 w\right)^{-1/2}\cdot \frac {p_y}{8}\right]$$

For this derivative to be negative, we want to expression inside the brackets to be positive. So we require

$$1- \left((p_y/8)^2 + \frac14 w\right)^{-1/2}\cdot \frac {p_y}{8} > 0$$

$$\implies 1> \left((p_y/8)^2 + \frac14 w\right)^{-1/2}\cdot \frac {p_y}{8} $$

$$ \implies \left((p_y/8)^2 + \frac14 w\right)^{1/2} > \frac {p_y}{8}$$

$$\implies (p_y/8)^2 + \frac14 w > (p_y/8)^2$$

which holds always. So we see that even this "strange looking" utiity function with its "messy" demand functions, reflects always the negative effect of price on quantity demanded.


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