On the graph below, R is the region of feasible solutions defined by inequalities y > 2, y = x + 1 and 5y + 8x < 92. Find the greatest value of 2y + x which satisfies the set of inequalities, where x and y are integers.
We are looking for integer values of x and y in the region R where 2y + x has the greatest value. We could substitute all the possible (x , y) values in R into 2y + x to get the largest value but that would be too long and tedious.
A better method would be to find the line 2y + x = c where x and y are in R and c has the largest possible value. In this case, the equation 2y + x = c is known as the linear objective function.
Rewriting 2y + x = c as y = – x + c, we find that the gradient of the line is – . We need to find a line with gradient – , within the region R that has the greatest value for c.
Draw a line on the graph with gradient – . (Any line with a gradient of – would be acceptable).
To look for the line, within R , with gradient – and the greatest value for c, we need to find the line parallel to the line drawn above that has the greatest value for c (the y-intercept). We can use the technique in the previous section to construct parallel lines. We will draw parallel lines with increasing values of c. (Increasing values of c means we move upwards). We will stop at the parallel line with the largest c that has the last integer value of (x , y) in the region R.
Now, we have all the steps that we need for solving linear programming problems, which are:
Step 1: Interpret the given situations or constraints into inequalities.
Step 2: Plot the inequalities graphically and identify the feasible region.
Step 3: Determine the gradient for the line representing the solution (the linear objective function).
Step 4: Construct parallel lines within the feasible region to find the solution.
Joanne wants to buy x oranges and y peaches from the store. She must buy at least 5 oranges and the number of oranges must be less than twice the number of peaches. An orange weighs 150 grams and a peach weighs 100 grams. Joanne can carry not more than 3.6 kg of fruits home.a) Write 3 inequalities to represent the information given above.
Solution:a) at least 5 oranges: x ≥ 5
c) We need to find the maximum that Joanne can spend buying the fruits. This would mean looking for the maximum value of c for 70x + 90y = c .
70x + 90y = c
We need to find the line with gradient with maximum value of c such that (x, y) is in the region S
Plot a line and with gradient move it to find the maximum within the region S. Draw parallel lines with increasing values of c. (Increasing values of c means we move upwards). Stop at the parallel line with the largest c that has the last integer value of (x , y) in the region S.
The maximum value is found at (5,28) i.e. 5 oranges and 28 peaches. Therefore, the maximum that Joanne can spend on the fruits is: 70 × 5 + 90 × 28 = 2870 cents = $28.70.
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