Question 14.6.1: Solve the consumer’s demand problem max U(x, y, z) = x²y³z ......
Solve the consumer’s demand problem
\max U(x, y, z) = x²y³z s.t. x + y + z = 12
With \mathcal{L}(x,y,z)=x^{2}y^{3}z-\lambda(x+y+z-12), the first-order conditions are
{\mathcal{L}}_{1}^{\prime}=2x y^{3}z-\lambda=0,\ {\mathcal{L}}_{2}^{\prime}=3x^{2}y^{2}z-\lambda=0,\enspace{\mathrm{and}}\enspace{\mathcal{L}}_{3}^{\prime}=x^{2}y^{3}-\lambda=0 (∗)
If any of the variables x, y, and z is 0, then x²y³z = 0, which is not the maximum value. So suppose that x, y, and z are all positive. From the two first equations in (∗), we have 2xy³z =3x²y²z, so y = 3x/2. The first and third equations in (∗) likewise imply that z = x/2. Inserting y = 3x/2 and z = x/2 into the constraint yields x + 3x/2 + x/2 = 12, so x = 4. Then y = 6 and z = 2. Thus, the only possible solution is (x, y, z) = (4, 6, 2).