The charges for hiring a car from three different companies, based on the number of days for which a car is hired and/or number of kilometres for which the car is driven are as follows:
Company A charges $660 per day
Company B charges 48 cents on per kilometre driven
Company C charges $300 per day and $150 on per kilometre driven
John needs to hire a car for 4 days to drive 560 kilometres
(i) Write down two matrices only such that the elements of their product under matrix multiplication give the charge of hiring a car from three different companies
(ii) Find this product. Hence, state the company that John should hire the car in order to save cost.
My answer:
The number of days - x; number of kilometres driven - y
Thus, (660x) =
(48y) =
(300x +150y)
This above is supposed to be a 3x1 matrix
I would equate all of these matrices to the 2x1 matrix (4x) (5y)
Please tell me if I'm on the right track. Thank you.
You are on the right track, but you made a small mistake in interpreting the given data. Let's correct it and find the matrices.
Let x be the number of days (4 days) and y be the number of kilometers driven (560 km).
The charges of each company can be represented as follows:
Company A: 660x
Company B: 48y
Company C: 300x + 150y
The 3x1 matrix for the charges would look like this:
| 660x |
C = | 48y |
| 300x+150y |
Now, we need to find the matrix that when multiplied by C gives the charges of the three companies. That matrix should have two elements: x (number of days) and y (number of kilometers driven). So, the 2x1 matrix would look like this:
| 4 |
D = |560 |
Now let's find the product of these two matrices to find the charges from each company:
E = CD
| 660*4 | | 4 | | 2640 |
E = | 48*560 | * |560 | = | 26880 |
| 300*4+150*560 | | 97800 |
The matrix E now contains the charges for each company:
Company A: $2640
Company B: $26880
Company C: $97800
To save cost, John should hire the car from Company A.
Well, you're on the right track in terms of setting up the matrices, but you have a little mistake in your equation.
Let's call the 2x1 matrix you're trying to find as M: (4x, 5y)
The charges for each company can be written as follows:
Company A: (660x, 0y)
Company B: (0x, 0.48y)
Company C: (300x, 150y)
Now, if we multiply these matrices:
M = (660x, 0y) + (0x, 0.48y) + (300x, 150y)
Performing the matrix addition:
M = (660x + 0x + 300x, 0y + 0.48y + 150y)
Simplifying:
M = (960x, 150.48y)
So, the product of the matrices gives us the charge for hiring a car from each company.
Now, let's see which company John should choose to save costs. We know that John needs to hire a car for 4 days (x = 4) and drive 560 km (y = 560).
Plugging these values into the matrix M, we get:
M = (960 * 4, 150.48 * 560)
= (3840, 84028.8)
So, the costs for hiring a car from each company are:
Company A: $3840
Company B: $84028.8
Company C: $3840 + $84028.8 = $87868.8
Since Company A charges the lowest amount, John should hire the car from Company A in order to save costs.
Hope that helps!
You are on the right track, but there are a few corrections that need to be made.
Let's define the matrices as follows:
Matrix A = [660x]
Matrix B = [0.48y]
Matrix C = [300x + 150y]
Now, we need to find the product of these matrices, which represents the charges from each company.
Matrix A * Matrix B = [660x] * [0.48y] = 0.48 * 660xy
Matrix A * Matrix C = [660x] * [300x + 150y] = 660 * (300x^2 + 150xy)
Matrix B * Matrix C = [0.48y] * [300x + 150y] = 0.48 * (300xy + 150y^2)
Now, we need to write down a 3x1 matrix [4x 560y], representing the number of days and the number of kilometers John wants to hire the car for.
To calculate the total charges from each company, we multiply this matrix by each of the previous matrices:
[4x 560y] * (Matrix A * Matrix B) = 0.48 * 660 * 4x * 560y
[4x 560y] * (Matrix A * Matrix C) = 660 * 300 * 4x * (4x + 150y)
[4x 560y] * (Matrix B * Matrix C) = 0.48 * 300 * (560y * 4x + 560y^2)
Now, you can simplify these expressions and compare the total charges from each company to determine which one John should hire the car from in order to save costs.
Yes, you're on the right track! Let's break it down further.
(i) To write down two matrices such that the elements of their product give the charges of hiring a car from three different companies, you can represent the charges for Company A, Company B, and Company C separately in two matrices.
First matrix, let's call it matrix A:
|660x|
|___ |
|___ |
This matrix represents the charge from Company A, where x is the number of days John wants to hire the car.
Second matrix, let's call it matrix B:
| 48y |
| ____ |
| ____ |
This matrix represents the charge from Company B, where y is the number of kilometers John wants to drive.
(ii) Now, you need to find the product of these two matrices. To do so, perform matrix multiplication where the number of columns in the first matrix equals the number of rows in the second matrix. Since both matrices have only one column, the product matrix will be a 1x1 matrix, representing the total cost.
Multiplying A and B, you get:
product matrix = A * B = |660x| * | 48y |
|____| | ____ |
= (660x) * (48y)
To simplify the expression, you can calculate the product of 660x and 48y, which gives:
product matrix = 31,680xy
The product matrix 31,680xy represents the total cost of hiring a car from three different companies.
To find out which company John should choose to save costs, we need to compare the charges from Company A, Company B, and Company C. Based on the given charges, the company with the lowest total cost is the most cost-effective option for John.