Optimization Problems

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Optimization goes this way: if it is loss then it should be minimum and if it is profit then it should be maximum. Different methods are used to solve these problems. One of the method is graphical method in which problem is first formulated and then it is graphed and a common region is found which is called feasible region in which we find boundary points which we put in optimization function. Some of the example problems to practice optimization are given as follows.

Linear Programming Optimization:

A general linear programming problem consists of maximizing or minimizing an objective function subject to certain given constraints.

Working rules for linear constraint optimization:

Step1: Identify the unknowns in the given LPP. Denote them by x and y.

Step2: Formulate the objective function in terms of x and y. Be sure whether it is to be maximized or minimized.

Step3: Translate all the constraints in the form of linear inequalities.

Step4: Solve these inequalities simultaneously. Mark the common area by shaded region. This is the feasible region.

Step5: Find the coordinates ...
... of all the vertices of the feasible region.

Step6: Find the value of the objective function at each vertex of the feasible.

Step7: Find the values of x and y for which the objective function z = ax + by has maximum or minimum value (as the case may be).

Example Problems:

Example 1: A firm plans to purchase at least 200 quintals of scrap containing high quality metal X and low quality metal Y. It decides that the scrap to be purchased must contain at least 100 quintal of X-metal and not more than 35 quintals of Y-metal. The firm can purchase the scrap form two suppliers (A and B) in unlimited quantities. The percentage of X and Y metals in terms of weight in the scraps supplied by A and B is given below:

Metals Supplier A Supplier B

X 25% 75%

Y 10% 20%

The price of A’s scrap is Rs 200 per quintal and that of B’s is Rs 400 per quintal. Formulate this problem as LP model and solve it to determine the quantities that the firm should buy from the two suppliers so as to minimize total purchase cost.

Solution: The formulation of the given problem is:

Min. (total cost) Z = 200 x1 + 400 x2,

Problem 1

Subject to the constraints:

x1 + x2 200, $\frac{1}{4}$x1 + $\frac{3}{4}$x2 100, $\frac{1}{10}$x1 + $\frac{1}{5}$x2 35, x1 0, x2 0.

Where x1, x2 represent the number of quintals of scrap for two suppliers A and B respectively.

The feasible region is the shaded area PQR which is obtained by drawing the graph of the constraints:

x1 + x2 200, x2 + 3x2 400 and x1 + 2x2 175

The coordinates of the corner points of the feasible region are:

P(100, 100), Q(50, 150), R(250, 50)

The Z has the min. value at the point P(100, 100). Thus the answer is x1 = 100, x2 = 100, min. Z = Rs. 60,000.

Example 2: Find the maximum and minimum values of 5x + 2y subject to the constraints

-2x – 3y = -6, i.e., 2x + 3y = 6; x – 2y = 2;

6x + 4y = 24; -3x + 2y = 3;

x = and y = 0.

Solution: 2x + 3y = 6

x/3 + y/2 = 1

This line meets the axes at (3,0) and (0,2). Join these points to obtain the line 2x + 3y = 6.

x/2 + y/-1 = 1

This line meets the axes at (2,0) and (0,-1). Join these points to obtain the line x – 2y = 2.

6x + 4y = 24

x/4 + y/6 = 1.

This line meets the axes at (4,0) and (0,6). Join these points to obtain the line 6x + 4y = 24.

-3x + 2y = 3

x/-1 + y/3,2 = 1.

This line meets the axes at (-1,0) and (0,3/2). Join these points to obtain the line -3x + 2y = 3.

Also, x = 0 is the y-axis and y = 0 is the x-axis.

Thus, all these lines include the quadrilateral ABCD, shown by the shaded region.

Thus shaded region represents all possible solutions of the system. Now, we may use the following two methods for obtaining the maximum and minimum values of 5x + 2y.

Check this Current Density awesome i recently used to see.

linear constraint optimization

Example 3: Solve for x and y for the optimization calculus problem:

`y =13x^2 + 18x`

Solution:

Step 1: given problem `y = 13x^2+ 18x`

Step 2: Differentiate with respect to x

`dy / dx` = 26x + 18

Equate `dy /dx ` to 0.

`dy / dx` = 26x + 18 = 0

26x = -18

x = `(-18) / (26)` or -0.69

Step 3: Insert x = 0.69 in the given equation

y = `13(-0.69) ^2+ 18(-0.69)`

= 13(0.4761) - (12.42)

= 6.18 - 12.42

= -6.24

Therefore, x = -0.69 and y = -6.24

Step 4: Using the given equation obtain the graph and plot the point in the graph.

Graph for optimization calculus problem

Example 4: The standard weight of a special purpose brick is 5 kg and it contains tow basic ingredients B1 ad B2. B1 costs Rs. 5 per kg and B2costs Rs. 8 per kg. strength considerations state that the brick contains not more than 4 kg of B1 and minimum of 2 kg of B2. Since the demand for the product is likely to be related to the price of the brick, find out graphically minimum cost of the brick satisfying the above conditions.

Solution: The formulation of the given problem is:

Min (total cost) Z = 5x1 + 8x2, subject to the constraints

problem(b)

x1 4, x2 2 and x1 + x2 = 5, x1 0, x2 0, where (x1 x2) = the amount of ingredients B1 (in kg.) respectively. The given constraints are plotted on the graph as shown in the figure. It may be observed that feasible region has two corner points P(3, 2) and Q(4, 2). The minimum value of Z is found at P(3, 2), i,e., x1 = 3, x2 = 2. Hence the optimum product B2 of a special case brick in order to achieve the minimum cost of Rs 31.

Example 5: Assume that the following specify a generalized linear programming problem:

f (x) = 2x1 +x2

subject to x1 + x2 6, x1 3, 2x1 + x2 4, x1, x2 0.

Solution: Graph this problem, identifying the three constraint equation lines and the feasible zone common to all of them. Plot dotted lines for values of 3. 6. 9 and 12 for the objective function f (x) and for what values of x1 and x2 does it occur.

problem(c)

The whole shaded region would lie in the 1st quadrant

problem 2

A = (2, 0), B = (3, 0) C = (3, 3),D = (6, 0), E = (4, 0)

f (A) = 2 × 2 = 1·0 = 4, f (B) = 2 × 3 + 1· 0 = 6, f (C) = 2 × 3 + 1 × 3 = 9, f (D) = 2 × 4 + 1 × 0 = 8

... f (C) = 9 is maximum for x1 = 3, x2 = 3

We had practised some optimization problems.

Learn more on about Area of a Rectangular Prism and its Examples. Between, if you have problem on these topics how to solve complex rational expressions, Please share your comments.