The problem asks us to find the values of the coefficients \( a \) and \( b \) in the objective function \( z = ax + by \). We are given that the maximum value of this objective function is 50, and this maximum occurs at two specific corner points of the feasible region: (0,10) and (10,6).

Understanding Linear Programming Corner Point Theorem

In Linear Programming, the maximum or minimum value of an objective function, if it exists within a bounded feasible region, always occurs at one of the corner points (vertices) of the feasible region. When the maximum or minimum value occurs at two distinct corner points, it means that all points on the line segment connecting these two corner points also yield the same maximum or minimum value.

Since the maximum value of \( z = ax + by \) is 50 and it occurs at both (0,10) and (10,6), we can use this information to set up equations for \( a \) and \( b \).

Setting Up Equations for Coefficients \( a \) and \( b \)

We are given the objective function \( z = ax + by \) and that the maximum value is 50 at two points:

• Point 1: \((x, y) = (0, 10)\)
• Point 2: \((x, y) = (10, 6)\)

Substituting these points into the objective function \( z = ax + by \) and setting \( z = 50 \), we get two linear equations:

For point (0, 10):

\( a(0) + b(10) = 50 \)

\( 10b = 50 \)   (Equation 1)

For point (10, 6):

\( a(10) + b(6) = 50 \)

\( 10a + 6b = 50 \)    (Equation 2)

Solving the System of Linear Equations

Now we have a system of two linear equations with two variables \( a \) and \( b \):

1) \( 10b = 50 \)

2) \( 10a + 6b = 50 \)

Let's solve Equation 1 for \( b \):

\( 10b = 50 \)

Divide both sides by 10:

\( b = \frac{50}{10} \)

\( b = 5 \)

Now substitute the value of \( b = 5 \) into Equation 2:

\( 10a + 6(5) = 50 \)

\( 10a + 30 = 50 \)

Subtract 30 from both sides of the equation:

\( 10a = 50 - 30 \)

\( 10a = 20 \)

Divide both sides by 10:

\( a = \frac{20}{10} \)

\( a = 2 \)

So, the values of \( a \) and \( b \) are \( a = 2 \) and \( b = 5 \).

Verification

Let's check if \( z = 2x + 5y \) gives a maximum value of 50 at the two given points:

• At (0, 10): \( z = 2(0) + 5(10) = 0 + 50 = 50 \). This is correct.
• At (10, 6): \( z = 2(10) + 5(6) = 20 + 30 = 50 \). This is correct.

The problem also lists other corner points: (4,2) and (3,7). The value of z should be less than or equal to 50 at these points for 50 to be the maximum value:

• At (4, 2): \( z = 2(4) + 5(2) = 8 + 10 = 18 \). \( 18 \le 50 \).
• At (3, 7): \( z = 2(3) + 5(7) = 6 + 35 = 41 \). \( 41 \le 50 \).

This verification confirms that with \( a = 2 \) and \( b = 5 \), the maximum value of 50 occurs at (0,10) and (10,6), and is less at the other given points.

Final Values

The value of \( a \) is 2 and the value of \( b \) is 5.

Revision Table: Key Concepts

Additional Information: LP Problem Structure

A standard Linear Programming problem involves:

1. An Objective Function: This is the function you want to maximize or minimize (e.g., profit, cost). It's typically in the form \( Z = c_1x_1 + c_2x_2 + \dots + c_nx_n \).
2. Decision Variables: These are the variables (like \( x \) and \( y \) in our problem) that you can control and whose values you are trying to find to optimize the objective function.
3. Constraints: These are linear inequalities or equalities that the decision variables must satisfy. They define the boundaries of the feasible region (e.g., resource limitations, production requirements). Constraints often include non-negativity constraints like \( x \ge 0, y \ge 0 \).
4. Feasible Region: The area on a graph that contains all points satisfying all constraints simultaneously. For two variables, this is a polygon (bounded) or an unbounded region.

Solving an LP problem often involves graphing the constraints to find the feasible region and then evaluating the objective function at each corner point of the feasible region to find the maximum or minimum value.