To maximize a linear objective function:
1. Identify all vertices of the feasible region.
2. Substitute the coordinates of each vertex into the objective function.
3. Choose the highest value for maximization.
Step 1: The objective function \( z = 5x + 2y \) is maximized at one of the vertices of the feasible region \( R_1 \). The vertices of \( R_1 \) are \( A(0, 50) \), \( B(20, 40) \), and \( C(50, 100) \).
Step 2: Evaluate \( z \) at each vertex: - At \( A(0, 50) \): \[ z = 5(0) + 2(50) = 100. \] - At \( B(20, 40) \): \[ z = 5(20) + 2(40) = 100 + 80 = 180. \] - At \( C(50, 100) \): \[ z = 5(50) + 2(100) = 250 + 200 = 450. \]
Step 3: Conclusion: The maximum value of \( z = 5x + 2y \) is 450 at \( C(50, 100) \).
Fit a straight-line trend by the method of least squares for the following data:
\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline \textbf{Year} & 2004 & 2005 & 2006 & 2007 & 2008 & 2009 & 2010 \\ \hline \textbf{Profit (₹ 000)} & 114 & 130 & 126 & 144 & 138 & 156 & 164 \\ \hline \end{array} \]When observed over a long period of time, a time series data can predict trends that can forecast increase, decrease, or stagnation of a variable under consideration. The table below shows the sale of an item in a district during 1996–2001:
\[ \begin{array}{|c|c|c|c|c|c|c|} \hline \textbf{Year} & 1996 & 1997 & 1998 & 1999 & 2000 & 2001 \\ \hline \textbf{Sales (in lakh ₹)} & 6.5 & 5.3 & 4.3 & 6.1 & 5.6 & 7.8 \\ \hline \end{array} \]