Answer to Question #102136 in Operations Research for Batool Amin

Question

V = xy+ λ(2,000-20x-10y)
where λ is the Lagrange multiplier.
Now, the first-order conditions for constrained output maximisation are
how i slove it

Answer 1

$\nabla_{x,y,\lambda} =(\frac{\partial V}{\partial x},\frac{\partial V}{\partial y}, \frac{\partial V}{\partial x}) =0$

Take partial derivatives of $V(x,y,\lambda)= xy+\lambda(2,000-20x-10y)$

$\frac{\partial V(x,t,\lambda)}{\partial x} = y-20\lambda$

$\frac{\partial V(x,t,\lambda)}{\partial y} = x-10\lambda$

$\frac{\partial V(x,t,\lambda)}{\partial \lambda} = 2,000-20x-10y$

Finally, we get the system of equations

$\begin{cases} y-20\lambda = 0 \\ x-10 \lambda=0\\ 2,000-20x-10y=0\end{cases}$

Solve this system:

$\begin{cases} y=20\lambda \\ x=10 \lambda\\ 2,000-200\lambda-200\lambda=0\end{cases}$

$\begin{cases} y=20\lambda \\ x=10 \lambda\\ 400\lambda= 2,000\end{cases}$

$\begin{cases} \lambda= 5 \\ x=50 \\y=100\end{cases}$