Matrices
The U-Derive Rent a Truck Company plans to spend $5 million on 300 new vehicles. Each van will cost $15,000 and each truck $25,000. Past experience shows that U-Drive needs twice as many as trucks. How many of each kind of vehicle can the company buy?
this is calculus?
t = 2v
t+v = 300
so, 200 trucks, 100 vans
25t + 15v <= 500
25*200 + 15*100 = 6500
So they can't buy all 300 vehicles.
So, what adjustment can you make?
v+t </= 300
v = 2 t
c = 15v + 25 t </= 5,000
3 t </= 300 so t</= 100
30 t + 25 t = 55 t </= 5,000
t </= 90.9
so buy 90 trucks
that costs 90 * 25,000 = 2,250,000
you have 5,000,000 - 2,250,000 = 2,750,000 left
2,750,000/15,000 = 183 vans
total of 273 vehicles
for
183*15,000 + 90*25,000 = $4,995,000
@oobleck thx for the response and yes we are learning this in Calculus and it suppose to be a project. i have no idea why it was labeled in Matrices section
@Damon thx for the answer
If you label your question as "Calculus" you only get crusty old mathematicians and retired scientists to read it or reply. You really needed someone who rents trucks out.
To solve this problem, we can use matrices and systems of equations. Let's denote the number of vans as "v" and the number of trucks as "t."
We are given the following information:
1. There will be a total of 300 new vehicles: v + t = 300
2. Each van costs $15,000 and each truck costs $25,000: 15,000v + 25,000t = 5,000,000
We can write these two equations as a system of equations in matrix form:
|1 1| |v| |300|
|15,000 25,000| |t| |5,000,000|
To solve for v and t, we need to find the inverse of the coefficient matrix and multiply it by the constant matrix:
|v| |1 1|^(-1) |300|
|t| = |15,000 25,000| |5,000,000|
Calculating the inverse of the coefficient matrix:
|1 1|^(-1) |25,000 -1|
|15,000 25,000| |-15,000 1|
Multiplying the inverse matrix by the constant matrix:
|v| |25,000 -1| |300|
|t| = |-15,000 1| |5,000,000|
Multiplying the matrices:
|v| |(25,000 * 300) + (-1 * 5,000,000)|
|t| = |(-15,000 * 300) + (1 * 5,000,000)|
Calculating the values:
|v| |2,500,000 - 5,000,000|
|t| = |-4,500,000 + 5,000,000|
|v| |-2,500,000|
|t| = |500,000|
Therefore, the company can buy -2,500,000 vans and 500,000 trucks. However, we cannot have negative numbers of vehicles, so the number of vans and trucks must be non-negative.
Since negative numbers of vans are not possible, the company cannot buy any vans. However, they can buy 500,000 trucks.