Link BD consists of a single bar 30 mm wide and 12 mm thick. Knowing that each pin has a10 mm diameter, determine the maximum value of the average normal stress in the link BD if θ = 15 degrees.

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Question:

A company produces three products, each of which must be processed through one department. Given table summerizes the labor hour and raw material requirements per unit of each product. Each month there are 1500 labor hours and 3800 pounds of raw materials available. If combined monthly production for the three products should equal 500 units, Determine a combination of three products which satisfies the given values.

 

		Products
	1	2	3
labor hour per unit	3	2	4
Pounds of raw materials /unit	10	8	6

A company produces three products, each of which must be processed through one department. Given table summerizes the labor hour and raw material requirements per unit of each product. Each month there are 1500 labor hours and 3800 pounds of raw materials available. If combined monthly production for the three products should equal 500 units, Determine a combination of three products which satisfies the given values.     Products    1 2 3labor hour per unit 3 2 4Pounds of raw materials /unit 10 8 6

Step 1

In linear programming problems, we have to convert our questions into suitable equations and solve these equations to get the solution values for the variables. Since there is no maximum or minimum condition, this question can be solved directly without much complexity. 

Step 2

Let us take the number of units produced for products one, two, and three as x,y, and z.

Then considering labour hour per unit and pounds of raw material per unit we can make two equations taking number units as x, y, and z.

First equation from labou hour/unit is 3x +2y +4z = 1500 and the second equation from pounds of raw material per unit is 10x +8y + 6z = 3800. 

Also considering the total number of units from three products we get x +y +z = 500.

x +y +z =500, from this we can write z = 500-(x+y)

Substituting this value in the first equation we get 3x +2y +4(500 -(x+y) = 1500

3x +2y + 2000-4x -4y =1500, this gives -x - 2y = -500, x +2y = 500

Substituting the value of z in second equation  we get 10x + 8y + 6( 500 -(x+y)) = 3800

10x +8y +3000 -6x -6y = 3800, this gives 4x +2y = 800

 

 

 

 

Step 3

Now we have to solve x +2y =500 and 4x +2y =800 to get the values of x and y.

x + 2y = 500, x =500 -2y and substitute this value in 4x +2y = 800

4(500 -2y) +2y = 800, this gives 2000 -8y + 2y = 800

-6y = -1200, y = -1200/-6 = 200

Now substituting the value of y in 4x + 2y = 800, we get 4x + 2*200 = 800, 4x = 800-400 =400

4x = 400, x =400/4 =100

We know z = 500 -(x+y) = 500 -(100 +200) = 200.

 

Step 4

So the final solution is x = 100 units, y = 200 units, and z = 200 units.