A civil engineer involved in construction requires 5000, 6000, 6200 and 5462 m3
ID: 2941693 • Letter: A
Question
A civil engineer involved in construction requires 5000, 6000, 6200 and 5462 m3 of sand, find gravel, coarse gravel and additives respectively, for a building project. There are four pits from which these materials can be obtained. The composition of these pits is:
How many cubic meters must be hauled from Pit 3 in order to meet the engineer`s needs?
(Use any numerical methods-Any unknowns finding technique like gauss-seidel, LU decomposition and such)
A civil engineer involved in construction requires 5000, 6000, 6200 and 5462 m3 of sand, find gravel, coarse gravel and additives respectively, for a building project. There are four pits from which these materials can be obtained. The composition of these pits is: How many cubic meters must be hauled from Pit 3 in order to meet the engineer's needs? (Use any numerical methods-Any unknowns finding technique like gauss-seidel, LU decomposition and such)Explanation / Answer
Misread 35 as 55 so I kept getting negative answers so I didn't use gauss originally. this should be accurate
Rewrite the matrix as its transpose and set it equal to 5000, 6000, 6200, and 5462 respectively
50 20 20 10 5000
30 40 10 15 6000
10 20 40 55 6200
10 20 30 40 5462
Divide row1 by 50
1 2/5 2/5 1/5 100
30 40 10 15 6000
10 20 40 35 6200
10 20 30 40 5462
Add (-30 * row1) to row2
1 2/5 2/5 1/5 100
0 28 -2 9 3000
10 20 40 35 6200
10 20 30 40 5462
Add (-10 * row1) to row3
1 2/5 2/5 1/5 100
0 28 -2 9 3000
0 16 36 33 5200
10 20 30 40 5462
Add (-10 * row1) to row4
1 2/5 2/5 1/5 100
0 28 -2 9 3000
0 16 36 33 5200
0 16 26 38 4462
Divide row2 by 28
1 2/5 2/5 1/5 100
0 1 -1/14 9/28 750/7
0 16 36 33 5200
0 16 26 38 4462
Add (-16 * row2) to row3
1 2/5 2/5 1/5 100
0 1 -1/14 9/28 750/7
0 0 260/7 195/7 24400/7
0 16 26 38 4462
Add (-16 * row2) to row4
1 2/5 2/5 1/5 100
0 1 -1/14 9/28 750/7
0 0 260/7 195/7 24400/7
0 0 190/7 230/7 19234/7
Divide row3 by 260/7
1 2/5 2/5 1/5 100
0 1 -1/14 9/28 750/7
0 0 1 3/4 1220/13
0 0 190/7 230/7 19234/7
Add (-190/7 * row3) to row4
1 2/5 2/5 1/5 100
0 1 -1/14 9/28 750/7
0 0 1 3/4 1220/13
0 0 0 25/2 2606/13
Divide row4 by 25/2
1 2/5 2/5 1/5 100
0 1 -1/14 9/28 750/7
0 0 1 3/4 1220/13
0 0 0 1 5212/325
Add (-3/4 * row4) to row3
1 2/5 2/5 1/5 100
0 1 -1/14 9/28 750/7
0 0 1 0 26591/325
0 0 0 1 5212/325
Add (-9/28 * row4) to row2
1 2/5 2/5 1/5 100
0 1 -1/14 0 232023/2275
0 0 1 0 26591/325
0 0 0 1 5212/325
Add (-1/5 * row4) to row1
1 2/5 2/5 0 157288/1625
0 1 -1/14 0 232023/2275
0 0 1 0 26591/325
0 0 0 1 5212/325
Add (1/14 * row3) to row2
1 2/5 2/5 0 157288/1625
0 1 0 0 70091/650
0 0 1 0 26591/325
0 0 0 1 5212/325
Add (-2/5 * row3) to row1
1 2/5 0 0 104106/1625
0 1 0 0 70091/650
0 0 1 0 26591/325
0 0 0 1 5212/325
Add (-2/5 * row2) to row1
1 0 0 0 6803/325
0 1 0 0 70091/650
0 0 1 0 26591/325
0 0 0 1 5212/325
So in order to get the amount in Pit 3 we just take the third row which is 81.818 but this must be multiplied by a factor of becuase we solve in percentages:
8181.8 m3
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