Question

The corner points of the feasible region determined by the system of linear constraints are (0, 10), (5, 5), (15, 15), (0, 20). Let $z = px + qy$, where $p, q>0$. The relation between $p$ and $q$, so that the maximum $z$ occurs at both points (15, 15) and (0, 20) is

• A. $p = q$
• B. $p = 2q$
• C. $q = 2p$
• D. $q = 3p$

Hint:

Whenever a problem states that an objective function attains its optimal value at multiple specific points, simply evaluate the function at those points and equate the results. This is a very common and straightforward question type in linear programming.

Answer 1

The Correct Option is D

To solve the given problem, we need to analyze when the line defined by the equation \( z = px + qy \) achieves the maximum value simultaneously at the points (15, 15) and (0, 20). These points are given as part of the feasible region defined by the constraints of the problem.

The function \( z = px + qy \) represents a family of lines in the xy-plane, where \( p \) and \( q \) are positive constants. In a situation where \( z \) reaches its maximum at more than one point, those points should lie on the same line (i.e., the line defined by the function \( z \)).

We need to determine the relation between \( p \) and \( q \) such that the line passes through both points (15, 15) and (0, 20).

Let's substitute these points into the equation \( z = px + qy \):

1. Substitute the point \((15, 15)\):
   \( z = p \times 15 + q \times 15 = 15p + 15q \)
2. Substitute the point \((0, 20)\):
   \( z = p \times 0 + q \times 20 = 20q \)

Since both points give the same value of \( z \), we equate the expressions:

\( 15p + 15q = 20q \)

Simplifying the equation, we have:

\( 15p = 20q - 15q \)
\( 15p = 5q \)
\( p = \frac{5}{15}q = \frac{1}{3}q \)

This implies \( q = 3p \).

Thus, the relation between \( p \) and \( q \) that ensures the maximum occurs at both points (15, 15) and (0, 20) is \( q = 3p \).

Therefore, the correct answer is:

Correct Answer: \( q = 3p \)