Expenditure Minimisation Problem: Satiated Preferences

Consider the utility function u(x, y) = -(x-2)^2-(y-1)^2.
For a target utility \mu \leq 0, find the Hicksian demand function (x^h,y^h)(p_x, p_y, \mu), and the expenditure function e(p_x, p_y, \mu).

Minimize

p_x x + p_y y

S.t.

-(x-2)^2 - (y-1)^2 \geq \mu

where

\mu \leq 0, \quad p_x, p_y > 0

now, as we move the signs

(x-2)^2 + (y-1)^2 \leq -\mu

and

-\mu \geq 0

Now ,
when
\mu = 0

(x-2)^2 + (y-1)^2 \leq 0

and the feasible point is (2,1)
now , when
\mu < 0
for this we will use lagrangian

\mathcal{L} = p_x x + p_y y + \lambda \left( (x-2)^2 + (y-1)^2 + \mu \right)

Differentiating wrt x

p_x + 2\lambda(x-2) = 0 \implies x = 2 - \frac{p_x}{2\lambda}

Differentating wrt y

p_y + 2\lambda(y-1) = 0 \implies y = 1 - \frac{p_y}{2\lambda}

now
Using the constraint:

(x-2)^2 + (y-1)^2 = -\mu

Substituting the expressions for x and y:

\frac{p_x^2 + p_y^2}{4\lambda^2} = -\mu

Therefore, solving for \lambda:

\lambda = \frac{\sqrt{p_x^2 + p_y^2}}{2\sqrt{-\mu}}

Substituting back

x^h = 2 - \frac{p_x \sqrt{-\mu}}{\sqrt{p_x^2 + p_y^2}}, \quad y^h = 1 - \frac{p_y \sqrt{-\mu}}{\sqrt{p_x^2 + p_y^2}}

now, for corner solutions:
when
x^h < 0
then
putting x=0 in the constraint gives:

4 + (y-1)^2 = -\mu

solving for y^h:

y^h = 1 - \sqrt{-\mu - 4}

similarly ,
y^h < 0
then
Setting y=0 in the constraint gives:

(x-2)^2 + 1 = -\mu

solving for x^h:

x^h = 2 - \sqrt{-\mu - 1}

Hicksian Demand:

(X^h, Y^h)(P_x,P_y,\mu) = \begin{cases} \left\{(2, 1)\right\} & \text{if } \mu = 0 \\[12pt] \left\{\left( 2 - \dfrac{p_x \sqrt{-\mu}}{\sqrt{p_x^2 + p_y^2}}, \, 1 - \dfrac{p_y \sqrt{-\mu}}{\sqrt{p_x^2 + p_y^2}} \right)\right\} & \text{if } \mu < 0 \text{ and } x^h, y^h \geq 0 \\[12pt] \left\{\left( 0, \, 1 - \sqrt{-\mu - 4} \right)\right\} & \text{if } x^h < 0 \\[12pt] \left\{\left( 2 - \sqrt{-\mu - 1}, \, 0 \right)\right\} & \text{if } y^h < 0 \end{cases}

Now expenditure function:
e = p_x x^h + p_y y^h
Substitute x^h

p_x x^h = p_x \left( 2 - \frac{p_x \sqrt{-\mu}}{\sqrt{p_x^2 + p_y^2}} \right)

so

= 2p_x - \frac{p_x^2 \sqrt{-\mu}}{\sqrt{p_x^2 + p_y^2}}

Substitute y^h
Similarly for y^h:

p_y y^h = p_y \left( 1 - \frac{p_y \sqrt{-\mu}}{\sqrt{p_x^2 + p_y^2}} \right)

so

= p_y - \frac{p_y^2 \sqrt{-\mu}}{\sqrt{p_x^2 + p_y^2}}

Add Both Parts

e = 2p_x - \frac{p_x^2 \sqrt{-\mu}}{\sqrt{p_x^2 + p_y^2}} + p_y - \frac{p_y^2 \sqrt{-\mu}}{\sqrt{p_x^2 + p_y^2}}
= 2p_x + p_y - \left( \frac{p_x^2 + p_y^2}{\sqrt{p_x^2 + p_y^2}} \right) \sqrt{-\mu}

Final Answer

\boxed{e(p_x, p_y, \mu) = 2p_x + p_y - \sqrt{-\mu}\sqrt{p_x^2 + p_y^2}}