How do I derive the demand function from

$$U=y + 2\sqrt{x}$$

Currently I have $MRS = \frac{1}{\sqrt{x}} = \frac{P_x}{P_y}$, so, $P_y = P_x\sqrt{x}$

Using the budget line: $$I = xP_x + yP_y \implies I = xP_x +y(P_x\sqrt{x})$$

Which simplifies to $$x = \frac{2I + y^2 P_x \pm y\sqrt{Px(4I+y^2P_x)}}{2P_x}$$

I'm not sure if this is correct, should I have did $MRS = \frac{1}{\sqrt {x}} = \frac{P_x}{P_y}$, so, $P_x = \frac{P_y}{\sqrt{x}}$ instead and substitute $\frac{P_y}{\sqrt{x}}$ into budget line? How do I determine what to substitute into the budget line?


1 Answer 1


This is an example of quasilinear utility. In general, a quasilinear utility function takes the form $$u(x,y)=f(x)+y.$$ The perfect substitutes utility function, for example, is a special case of the quasilinear utility function. One feature of quasilinear utility is that the MRS is independent of at least one of the two goods: $$\frac{\frac{\partial u(x,y)}{\partial x}}{\frac{\partial u(x,y)}{\partial y}}=f'(x),$$ meaning the slope of indifference curves is independent of the quantity of good $y$.

The Marshallian demand functions satisfy the equations: $$ f'(x)=\frac{P_x}{P_y}$$ $$I=P_xx+P_yy,$$

which come from the first-order conditions of the constrained maximization problem. We can solve for the Marshallian demand function for $x$ directly from the first equation: $$x^*=f'^{-1}(\frac{P_x}{P_y}).$$ Substituting this into your second equation gives $$I=P_xf'^{-1}(\frac{P_x}{P_y})+P_yy$$ $$y^*=\frac{I-P_xf'^{-1}(\frac{P_x}{P_y})}{P_y}.$$

In your example, we get $$x^*=(\frac{P_y}{P_x})^2.$$ $$y^*=\frac{I-P_x(\frac{P_y}{P_x})^2}{P_y}.$$

There is a fundamental problem with your solution. You write a demand function for $x$ that depends on the quantity of $y$. Demand functions depend either on $(P_x,P_y,I)$, if Marshallian, or $(P_x,P_y,U)$, if Hicksian.


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