A manufacturer produces three types of radios: deluxe, standard and economy. Each radio uses three different types of transistors: P, Q and R. The deluxe radio uses 2 P's, 7 Q's and 1 R. The standard contains 2 P's, 3 Q's and 1 R, and the economy model requires 1 P, 2 Q's and 2 R's.
How many radios of each type can be constructed if the total number of transistors (P's, Q's and R's) available are 2200, 3400 and 1400 respectively and all transistors must be used?
Sorry about the confusion, which is not helped with at least 6 copies of identical questions, in addition to others that may have been deleted.
http://www.jiskha.com/display.cgi?id=1310358660
http://www.jiskha.com/display.cgi?id=1310104811
http://www.jiskha.com/display.cgi?id=1310104728
http://www.jiskha.com/display.cgi?id=1310355002
http://www.jiskha.com/display.cgi?id=1310104839
http://www.jiskha.com/display.cgi?id=1310413630
It would be much easier to respond to the original post, or make a new post with reference to the old post by number or by date and time.
Let
D=number of deluxe units
S=number of standard units
E=number of economy units
Set up equations for each type of transistor, forcing the total number used to the quantity available:
For type P:
2D+2S+E=2200
For type Q:
7D+3S+2E=3400
For type R:
D+S+2E=1400
Solving the above three simultaneous equations will give the quantities required. I get a total of 1200 units.
To solve this problem, we need to set up a system of equations to represent the given information.
Let's assume that the number of deluxe radios produced is D, the number of standard radios is S, and the number of economy radios is E.
According to the information given, each deluxe radio requires 2 P's, 7 Q's, and 1 R. Therefore, the total number of P transistors used in deluxe radios is 2D, the total number of Q transistors used is 7D, and the total number of R transistors used is D.
Similarly, for standard radios, the total number of P transistors used is 2S, the total number of Q transistors used is 3S, and the total number of R transistors used is S.
For economy radios, the total number of P transistors used is E, the total number of Q transistors used is 2E, and the total number of R transistors used is 2E.
Based on the above information, we can write the following equations:
2D + 2S + E = 2200 (equation 1)
7D + 3S + 2E = 3400 (equation 2)
D + S + 2E = 1400 (equation 3)
Now, we have a system of three equations with three variables. To solve this system, we can use various methods such as substitution, elimination, or matrices.
Let's use the substitution method to solve this system:
From equation 3, we can express D in terms of S and E:
D = 1400 - S - 2E
Substitute this expression into equations 1 and 2:
2(1400 - S - 2E) + 2S + E = 2200
7(1400 - S - 2E) + 3S + 2E = 3400
Simplify and solve these equations to find the values of S and E:
2800 - 2S - 4E + 2S + E = 2200
9800 - 7S - 14E + 3S + 2E = 3400
2800 - 3E = 2200
9800 - 4S - 12E = 3400
-3E = -600
-4S - 12E = -6400
Solving for E:
E = -600 / -3
E = 200
Substitute this value of E back into equation 1 to find D:
2D + 2S + 200 = 2200
2D + 2S = 2000
D + S = 1000
Since we know that D + S + 2E = 1400, we can substitute the values of S (from D + S = 1000) and E (from E = 200) into this equation to find the value of D:
D + 1000 + 2(200) = 1400
D + 1000 + 400 = 1400
D + 1400 = 1400
D = 0
Therefore, the solution to the system is D = 0, S = 1000, and E = 200.
This means that no deluxe radios can be produced, 1000 standard radios can be produced, and 200 economy radios can be produced if all transistors must be used.