\max_{x, y} \quad u(x,y) = -(x-2)^2 - (y-1)^2
\quad \text{s.t.} \quad p_x x + p_y y \leq M, \quad x, y \geq 0
The bliss point is:
(x_b, y_b) = (2, 1)
At this point, the maximum possible utility is:
u(2, 1) = 0
The cost of the bliss point is - 2p_x + p_y.
Now
Case 1
If M \geq 2p_x + p_y,
The optimal bundle is:
(x, y) = (2, 1)
With maximum utility:
u = 0
Case 2
If
M < 2p_x + p_y
bliss point is not affordable
now it should bind :
p_x x + p_y y = M
By using Lagrangian:
\mathcal{L} = -(x-2)^2 - (y-1)^2 + \lambda (M - p_x x - p_y y)
Differentiating w.r.t x
-2(x-2) - \lambda p_x = 0 \implies x = 2 - \frac{\lambda p_x}{2}
Differentiating w.r.t y
-2(y-1) - \lambda p_y = 0 \implies y = 1 - \frac{\lambda p_y}{2}
Substituting x and y
p_x \left( 2 - \frac{\lambda p_x}{2} \right) + p_y \left( 1 - \frac{\lambda p_y}{2} \right) = M
2p_x + p_y - \frac{\lambda}{2}(p_x^2 + p_y^2) = M
\therefore \lambda = \frac{2(2p_x + p_y - M)}{p_x^2 + p_y^2}
Again Substituting
x = 2 - \frac{p_x}{2} \cdot \frac{2(2p_x + p_y - M)}{p_x^2 + p_y^2} \implies x = 2 - \frac{p_x(2p_x + p_y - M)}{p_x^2 + p_y^2}
similarly
y = 1 - \frac{p_y(2p_x + p_y - M)}{p_x^2 + p_y^2}
Now checking corner points :
when x = 0
then y = M/Py
u = -4 - \left( \frac{M}{p_y} - 1 \right)^2
when y = 0
then x = M/Px
u = -\left( \frac{M}{p_x} - 2 \right)^2 - 1
Therefore , utility maximization is :
\boxed{
(X^d, Y^d)(P_x,P_y,M) =
\begin{cases}
\{(2, 1)\} & ,\text{if } M \geq 2p_x + p_y \\[12pt]
\left\{ \left( 2 - \dfrac{p_x(2p_x + p_y - M)}{p_x^2 + p_y^2}, \, 1 - \dfrac{p_y(2p_x + p_y - M)}{p_x^2 + p_y^2} \right) \right\} & \text{if } M < 2p_x + p_y \text{ and } x, y \geq 0 \\[12pt]
\left\{\left( 0, \dfrac{M}{p_y} \right)\right\} &, \text{if interior } x < 0 \\[12pt]
\left\{\left( \dfrac{M}{p_x}, 0 \right)\right\} &, \text{if interior } y < 0
\end{cases}
}