Answer to Question #113785 in Calculus for Naim

Question

A certain company manufactures a certain product, P , from three raw materials, combined according to a production function given by
2/5 1/5 1/5 P=50x y z .
Since the budget available for the purchase of the three raw materials of 24000 euros and knowing that they cost for unit 80, 12 and 10 euros, respectively, for x, y and z, what combination should you have regarding the purchase of products x, y and z, in order to maximize production?

Answer 1

$\phi(u)=-50x^{2/5}y^{1/5}z^{1/5}\\ \min \phi(u)\\ g_0(u)=80x+12y+10z-24000\\ g_0(u)\leq 0\\ g_1(u)=-x\\ g_1(u)<0\\ g_2(u)=-y\\ g_2(u)<0\\ g_3(u)=-z\\ g_3(u)<0\\ u=(x,y,z)\in \mathbb{R^3}$

$\frac{\partial g_0^T(u)}{\partial u}=(80\ 12\ 10)\\ \frac{\partial g_1^T(u)}{\partial u}=(-1\ 0\ 0)\\ \frac{\partial g_2^T(u)}{\partial u}=(0\ -1\ 0)\\ \frac{\partial g_3^T(u)}{\partial u}=(0\ 0\ -1)\\$

Only $g_0$ may be active.

$g_0\neq 0$

Hence each acceptable point is ordinary.

We will build normal (classical) Lagrange function:

$L(x,\lambda)=\phi(u)+\lambda_0 g_0(u)+\lambda_1 g_1(u)+\lambda_2 g_2(u)\\ +\lambda_3 g_3(u)=-50x^{2/5}y^{1/5}z^{1/5}+\lambda_0(80x+12y+\\ +10z-24000)-\lambda_1x-\lambda_2y-\lambda_3z.$

Due to necessary condition there exist $\lambda_0,\lambda_1,\lambda_2,\lambda_3\ (\lambda_i\geq 0)$ such that:

$\frac{\partial L(u,\lambda)}{\partial x}=-20x^{-3/5}y^{1/5}z^{1/5}+80\lambda_0-\lambda_1=0\\ \frac{\partial L(u,\lambda)}{\partial y}=-10x^{2/5}y^{-4/5}z^{1/5}+12\lambda_0-\lambda_2=0\\ \frac{\partial L(u,\lambda)}{\partial z}=-10x^{2/5}y^{1/5}z^{-4/5}+10\lambda_0-\lambda_3=0\\ \lambda_0(80x+12y+10z-24000)=0\\ \lambda_1=\lambda_2=\lambda_3=0.$

We have the system of equations:

$-20x^{-3/5}y^{1/5}z^{1/5}+80\lambda_0=0\\ -10x^{2/5}y^{-4/5}z^{1/5}+12\lambda_0=0\\ -10x^{2/5}y^{1/5}z^{-4/5}+10\lambda_0=0\\ \lambda_0(80x+12y+10z-24000)=0$

Using CoCalc (Sage) we find the solution of this system:

$u_0=(x_0,y_0,z_0)\\ x_0=150, y_0=500, z_0=600.$

Next we will check sufficient condition.

$\frac{\partial^2 L(u_0, \lambda^*)}{\partial u^2}$ will be a matrix of positive elements $a_{ij}$ with 3 columns and 3 rows.

$I=u^T \frac{\partial^2 L(u_0, \lambda^*)}{\partial u^2}u\\ u^T=(x\ y\ z)\\ I=a_{11}x^2+2a_{12}xy+2a_{13}xz+\\ +a_{22}y^2+2a_{23}yz+a_{33}z^2\\ \frac{\partial g_0^T(u_0)}{\partial u}u=0\\ (80\ 12\ 10)(x\ y\ z)^T=0\\ 80x+12y+10z=0\text{ --- our hyperplane}.\\ I>0 \text{ on this hyperplane}.$

So we have $x=150, y=500, z=600.$