Linear Programming- A shipment of three products in a cargo plane.

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Suppose we want to maximize revenue in the following situation.

We have 3 shipments of products.

Shipment one:

               *Has 10 ton of loads available for shipment  

                *A revenue of 700 euro per ton.   

              *Occupies 2000 units of Volume per ton

Shipment two:

              *Has 12 ton of loads available for shipment
             
              *A revenue of 725 euro per ton. 
             
             *Occupies 3500 units of Volume per ton

Shipment three:

              *Has 17 ton of loads available for shipment
             
               *A revenue of 685 euro per ton. 
               
               *Occupies 3000 units of Volume per ton

The plane which will carry these shipments can hold in total 32 ton

It is also split into 3 sections on the left and right:

Right Side

Hold A: Volume=16,000 units Weight=18% of total ( which I calculated to be 5.76 ton)

Hold B: Volume=20,000 units Weight=25% of total ( which I calculated to be 8 ton)

Hold C: Volume=14,000 units Weight=7% of total ( which I calculated to be 2.24 ton)

Left side

Hold A: Volume=10,000 units Weight=18% of total ( which I calculated to be 5.76 ton)

Hold B: Volume=20,000 units Weight=25% of total ( which I calculated to be 8 ton)

Hold C: Volume=12,000 units Weight=7% of total ( which I calculated to be 2.24 ton).

The Objective funtion

I believe would be

$$700x+725y+685z=R$$

Where $x=$ number of tons of shipment 1, $y=$number of tons of shipment 2, $z=$ number of tons of shipment 3

Constraints

I would have thought the constraints would be

  1. $$0 \leq x \leq 10$$, $$0\leq y\leq 12$$, $$0\leq z\leq17$$ ( as there is a limited amount of each shipment in tons)

  2. $$x+y+z\leq32$$ ( as the total amount of tons that can fit on the plane is 32)

  3. $$2000x+3500y+3000z\leq92,000$$ (As each ton of shipment occupies a specific volume and there is a max Volume that the combination of all the holds on the right and left can hold.

Then I thought separate constraints would be needed for each of the individual containers.

Note: I edited in the subscripts after the advice of a comment. These variables are such that

$x_{1a}+x_{1b}+x_{1c}+x_{2a}+x_{2b}+x_{2c}=x$

$y_{1a}+y_{1b}+y_{1c}+y_{2a}+y_{2b}+y_{2c}=y$

$z_{1a}+z_{1b}+z_{1c}+z_{2a}+z_{2b}+z_{2c}=z$

Right hold A

$$x_{1a}+y_{1a}+z_{1a}\leq 5.76$$

$$2000x_{1a}+3500y_{1a}+3000z{1a}\leq 16,000$$

Right hold B

$$x_{1b}+y_{1b}+z_{1b}\leq 8$$

$$2000x_{1b}+3500y_{1b}+3000z_{1b}\leq 20,000$$

Right hold C

$$x_{1c}+y_{1c}+z_{1c}\leq 2.24$$

$$2000x_{1c}+3500y_{1c}+3000z_{1c}\leq 14,000$$

Left hold A

$$x_{2a}+y_{2a}+z_{2a}\leq 5.76$$

$$2000x_{2a}+3500y_{2a}+3000z_{2a}\leq 10,000$$

Left hold B

$$x_{2b}+y_{2b}+z_{2b}\leq 8$$

$$2000x_{2b}+3500y_{2b}+3000z_{2b}\leq 20,000$$

Left hold C

$$x_{2c}+y_{2c}+z_{2c}\leq 2.24$$

$$2000x_{2c}+3500y_{2c}+3000z_{2c}\leq 12,000$$

After this I thought that there would be a maximum for the revenue for each hold and I could use this to find the overall maximum. However when I put this into Wolfram calculator I found that it says there are no local Maxima. Could someone help me see where I'm going wrong and advise me on what I should do instead ?

Many thanks in advance.