The optimal solution to the problem: Minimize : $f\left(X_{1}, X_{2}\right)=2 X_{1}^{2}-12 X_{1}+2 X_{1} X_{2}+X_{2}^{2}-4 X_{2}+160$ is
A) $X_{1}=2, X_{2}=0$
B) $X_{1}=4, X_{2}=-4$
C) $X_{1}=4, X_{2}=-2$
D) $X_{1}=-3, X_{2}=4$

Question

Quizplus · Accepted Answer

To find the optimal solution, we need to take the partial derivatives of $f(X_1, X_2)$ with respect to $X_1$ and $X_2$, set them to zero, and solve the resulting system of equations. The partial derivative with respect to $X_1$ is $4X_1 + 2X_2 - 12$, and with respect to $X_2$ is $2X_1 + 2X_2 - 4$. Setting these equal to zero gives us a system of equations:1. $4X_1 + 2X_2 = 12$2. $2X_1 + 2X_2 = 4$Solving this system yields $X_1 = 4$ and $X_2 = -2$, which corresponds to option C.

The Optimal Solution to the Problem: Minimize f(X1,X2)=2X12−12X1+2X1X2+X22−4X2+160f\left(X_{1}, X_{2}\right)=2 X_{1}^{2}-12 X_{1}+2 X_{1} X_{2}+X_{2}^{2}-4 X_{2}+160f(X1​,X2​)=2X12​−12X1​+2X1​X2​+X22​−4X2​+160

The Optimal Solution to the Problem: Minimize $f\left(X_{1}, X_{2}\right)=2 X_{1}^{2}-12 X_{1}+2 X_{1} X_{2}+X_{2}^{2}-4 X_{2}+160$