Company produces three models of speakers. models A,B,and C and can deliver them by truck, van and car. Truck holds 2 boxes of model A, 2 of model B and 3 of model C. Van holds 3 boxes of model A, 4 boxes of model B, and 2 boxes of model C The car holds 3 boxes of model A 5 boxes of model B and 1 box of model C

Question

Company produces three models of speakers. models A,B,and C and can deliver them by truck, van and car. Truck holds 2 boxes of model A, 2 of model B and 3 of model C. Van holds 3 boxes of model A, 4 boxes of model B, and 2 boxes of model C The car holds 3 boxes of model A 5 boxes of model B and 1 box of model C

If 25 boxes of model A, 33 boxes of model B and 22 boxes of model C are to be delivered how many vehicles of each type should be used so that all operate at full capacity. Part B of this question Model C has been discontinued. If 25 boxes of model A and 33 boxes of model B are to be delivered how many vehicles of each type should be used so that all operate at full capacity

Answer 1

I see the following matrix equation:

[t,v,c)
x
│2 2 3│
│3 4 2│
│3 5 1│
=
[25 33 22]

or
2t + 3v + 3c = 25
2t + 4v + 5c = 33
3t + 2v + c = 22

solve it as a system of equations.

Answer 2

I should be good to go from here. Just to old for this but I am determined to understand word problems. In my sixties

Answer 3

'In my sixties' ...

Hang in there, I am beyond that.

Answer 4

To determine the number of vehicles of each type needed to deliver the boxes at full capacity, we can set up a system of equations.

Let's solve Part A first:

Let t = number of trucks
Let v = number of vans
Let c = number of cars

The constraints are as follows:
2t + 3v + 3c = 25 (model A)
2t + 4v + 1c = 33 (model B)
3t + 2v = 22 (model C)

We need to find the values of t, v, and c that satisfy these equations and result in full capacity usage. To solve these equations, we can use substitution or elimination method.

Using the elimination method, we can eliminate model C from the equations by multiplying the third equation by 3, and then subtracting it from the first equation:

6t + 9v + 9c = 75
-(6t + 4v + 2c = 66)
-----------------------
5v + 7c = 9

Simplifying further, we get:

5v + 7c = 9 ... (i)
2t + 4v + 1c = 33 ... (ii)

Next, we can eliminate model C from the equations by multiplying the third equation by 2, and then subtracting it from the second equation:

2t + 4v + 1c = 33
-(6t + 4v + 2c = 66)
-----------------------
-4t - c = -33

Simplifying further, we get:

4t + c = 33 ... (iii)
-4t - c = -33 ... (iv)

By adding equations (iii) and (iv), we eliminate the variable c:

4t + c + (-4t - c) = 33 + (-33)
0 = 0

This result implies that the value of c is not determined by the equations. Therefore, there is an infinite number of solutions for c, and as long as the other variables satisfy the equations, the vehicles will operate at full capacity. Hence, for Part A, we cannot determine the exact number of vehicles of each type needed.

Now, let's solve Part B:

Since Model C has been discontinued, we can remove it from the equations and use only the equations for Models A and B:

2t + 3v = 25 ... (v)
2t + 4v = 33 ... (vi)

By subtracting equation (v) from equation (vi), we can eliminate t:

(2t + 4v) - (2t + 3v) = 33 - 25
v = 8

Substituting the value of v back into equation (v), we can solve for t:

2t + 3(8) = 25
2t + 24 = 25
2t = 1
t = 1/2

Since the number of vehicles cannot be fractional, we round t down to 0. Hence, for Part B, we would need 0 trucks, 8 vans, and an additional vehicle to carry the remaining 25 boxes of Model A. This additional vehicle can be a car.

To summarize:

Part A:
Trucks: Not determinable
Vans: Not determinable
Cars: Not determinable

Part B:
Trucks: 0
Vans: 8
Cars: 1