Manufacturing Company plans to spend $3 million on 200 new Technologies.
- Each Type-A Technologies will cost $10,000,
- Each Type-B Technologies will cost $15,000.
The company needs twice as many as Type-B Technologies.
How many of each kind of Technologies can the company buy?
* Solve The Question using Matrices
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To solve this question using matrices, we can set up a system of equations. Let's define the variables:
- Let x represent the number of Type-A technologies.
- Let y represent the number of Type-B technologies.
Given the information:
1. Each Type-A technology costs $10,000, which means the total cost of Type-A technologies is 10,000x.
2. Each Type-B technology costs $15,000, which means the total cost of Type-B technologies is 15,000y.
3. The company plans to spend $3 million, which means the total cost of technologies is 3,000,000.
Now we can set up the equations:
1. The total cost equation: 10,000x + 15,000y = 3,000,000
2. The equation stating that the company needs twice as many Type-B technologies as Type-A technologies: y = 2x
Now we can represent these equations in matrix form:
| 10000 15000 | | x | = | 3000000 |
| -1 2 | | y | | 0 |
To solve this, we can use matrix algebra. We need to find the inverse of the coefficient matrix:
| 10000 15000 |
| -1 2 |
Once you have found the inverse matrix, you can multiply it with the constant matrix [3000000, 0] to solve for x and y. The resulting values of x and y will give you the number of each kind of technologies the company can buy.
(Note: The final calculation is best done using a matrix calculator or software that supports matrix algebra.)