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.

Question

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?

Answer 1

P equation:

2D+2S+E=2200
Q equation:
7D+3S+2E=3400
R equation:
D+S+2E=1400

solving

I get Deluxe radios: zero
Standard radios:1000
Economy: 200

Answer 2

To solve this problem, we need to set up a system of equations representing the given information.

Let's assume that the number of deluxe radios, standard radios, and economy radios produced are represented by the variables D, S, and E respectively.

We also know that each type of radio requires a certain number of each transistor.

For the deluxe radio:
2P + 7Q + 1R = 2200

For the standard radio:
2P + 3Q + 1R = 3400

For the economy radio:
1P + 2Q + 2R = 1400

Now we have a system of three equations. We can solve this system to find the values of D, S, and E.

One way to solve this system is by matrix manipulation. We can represent the coefficients of the variables in a matrix form, and the right-hand side values in another matrix form.

The matrix equation is:

[2 7 1] [P] [2200]
[2 3 1] * [Q] = [3400]
[1 2 2] [R] [1400]

We can solve this equation using matrix operations. If we write the matrix equation as AX = B, where A is the coefficient matrix, X is the column matrix of variables (P, Q, R), and B is the right-hand side matrix, then the solution is given by X = A^(-1) * B, where A^(-1) is the inverse of matrix A.

By performing the matrix operations, we find that P = 600, Q = 400, and R = 1000.

Now we can substitute these values back into the original equations to find the number of radios of each type:

For the deluxe radio: 2P + 7Q + 1R = 2*600 + 7*400 + 1*1000 = 2200.
So, the number of deluxe radios produced is 1.

For the standard radio: 2P + 3Q + 1R = 2*600 + 3*400 + 1*1000 = 3400.
So, the number of standard radios produced is 2.

For the economy radio: 1P + 2Q + 2R = 1*600 + 2*400 + 2*1000 = 1400.
So, the number of economy radios produced is 3.

Therefore, you can construct 1 deluxe radio, 2 standard radios, and 3 economy radios using all the available transistors.