Question #102136
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
1
Expert's answer
2020-02-03T05:13:29-0500

x,y,λ=(Vx,Vy,Vx)=0\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,λ)=xy+λ(2,00020x10y)V(x,y,\lambda)= xy+\lambda(2,000-20x-10y)

V(x,t,λ)x=y20λ\frac{\partial V(x,t,\lambda)}{\partial x} = y-20\lambda

V(x,t,λ)y=x10λ\frac{\partial V(x,t,\lambda)}{\partial y} = x-10\lambda

V(x,t,λ)λ=2,00020x10y\frac{\partial V(x,t,\lambda)}{\partial \lambda} = 2,000-20x-10y

Finally, we get the system of equations

{y20λ=0x10λ=02,00020x10y=0\begin{cases} y-20\lambda = 0 \\ x-10 \lambda=0\\ 2,000-20x-10y=0\end{cases}

Solve this system:

{y=20λx=10λ2,000200λ200λ=0\begin{cases} y=20\lambda \\ x=10 \lambda\\ 2,000-200\lambda-200\lambda=0\end{cases}


{y=20λx=10λ400λ=2,000\begin{cases} y=20\lambda \\ x=10 \lambda\\ 400\lambda= 2,000\end{cases}


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





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