Question 15.22: An individual has the utility function U = 4X^0.5Y^0.5 and c......

The Lagrange function is

G = 4X^{0.5}Y^{0.5} + λ(100 − 2X − 8Y )

Differentiating and setting equal to zero to get the FOC for a maximum

G_X = 2X^{-0.5}Y^{0.5} − 2λ = 0         (1)
G_Y = 2X^{0.5}Y^{-0.5} − 8λ = 0          (2)
G_\lambda = 100 − 2X − −8Y = 0                        (3)

From (1)

X^{-0.5}Y^{0.5} = λ

From (2)

0.25X^{0.5}Y^{-0.5} = λ

Therefore

X^{-0.5}Y^{0.5} = 0.25X^{0.5}Y^{-0.5}

Multiplying both sides by 4X^{0.5}Y^{0.5}

4Y = X                 (4)

Substituting (4) into (3)

           100 − 2(4Y) − 8Y = 0
                                         Y = 6.25
and thus from (4)         X = 25

Differentiating (1), (2) and (3) again gives the bordered Hessian of second-order partial derivatives

H_{B} = \begin{bmatrix} U_{XX} & U_{XY} & −P_X \\ U_{YX}& U_{YY} &−P _Y \\−P_X& −P_Y &0 \end{bmatrix}  =  \begin{bmatrix} −X^{−1.5}Y^{0.5}& X^{−0.5}Y^{−0.5}& −2 \\ X^{−0.5}Y^{−0.5}& −X^{0.5}Y^{−1.5}& −8 \\ −2& −8& 0 \end{bmatrix}
=\begin{bmatrix}−0.02 & 0.08 &−2 \\ 0.08& −0.32 & −8 \\−2 &−8& 0 \end{bmatrix}

The determinant of this bordered Hessian, expanding along the third row is

\left|H_{B}\right| = −2(−0.64 − 0.64) + 8(0.16 + 0.16) = 2.56 + 2.56 = 5.12 > 0

and so the second-order conditions for a maximum are satisfied.