A total of 50000 is invested in three funds paying 6 8 and 1

A total of $50,000 is invested in three funds paying 6%, 8%, and 10% simple interest. The yearly interest is $3, 700. Twice as much money is invested at 6% as invested at 10%. How much was invested in each of the funds In an industrial process, water flows through three tanks in succession as illustrated in the figure. The tanks have unit cross-section and have heads of water x, y and z respectively. The rate of inflow into the first tank is u, the flow rate in the tube connecting tanks 1 and 2 is 6(x-y), the flow rate in the tube connecting tanks 2 and 3 is 5(y-z), and the rate of outflow from tank 3 is 5z. Assuming the system at steady stale, solve for the level of water (x, y, z) in each tank. lawnco produces three grades of commercial fertilizers. A 100-lb bag of grade-A fertilizer contains 18 lb of nitrogen, 4 lb of phosphate, and 5 lb of potassium. A 100-lb bag of grade-B fertilizer contains 20 lb of nitrogen and 4 lb each of phosphate and potassium. A 100-lb bag of grade-C fertilizer contains 24 lb of nitrogen, 3 lb of phosphate, and 6 lb of potassium. How mass 100-lb logs of each of the three grades of fertilizers should Lawnco produce if 26, 400 lb of nitrogen, 4900 lb of phosphate, and 6200 lb of potanium are available and all the nutrients are used? Provide the system of linear equation needed to find the values of the different traffic flows in matrix form. Solve for x_1, x_2, x_3 and x_4

Solution

Answer of part (a).

LEt three different amounts invested be $x,$y and $z respectively.

So by first condition, we have

x+y+z= 50000 ..............(i)

And using interest part, our second equation is :

0.06x+ 0.08y+0.1z= 3700 ................(ii)

And based on third condition total money invested at 6% is twice that of invested on 10%.

i.e. x= 2z ..........(iii)

So using (i) and (iii), we have

2z+y+z=50000 or 3z+y= 50000 or y= 50000 - 3z ...............(iv)

And using (ii) and (iii), we have (ii) ewquation as :

0.06(2z)+ 0.08(50000-3z) +0.1z = 3700

or 0.12z+4000-0.24z+0.1z = 3700

or -0.02z= 3700 - 4000 = - 300

\\\\z=300/0.02 = 15000

So x= 2(15000)= 30000

and y= 50000- 3z = 50000- 45000= 5000

So three amounts invested are $ 15000, $ 30000 and $5000.

Answer