Re: optimisation problem!!!

jib jib wrote:
> there is a clothing company that makes a up market line and a budget line of clothing using 3 materials:
> the up-market line = 50%A + 20%B + 30%C
> the budget line = 20%A + 50%B + 30%C
> market price are $790/tonne for upmarket and $700 for the budget line.
> For a period the company can by 100 metres of A for $800/metre, 100 metres of B for $600/metre and 75 metres of C for $400/metre.
>
> first you have to calculate the profit per meter for each line. and then find out how much of each line the company should produce to maximise it's profit.

Ok, first, $790/tonne is "up market?" Who buys "up market" clothing
by weight? (Or mass, if you want to be picky.)

Next, I don't think there's enough information here. Far as I can
tell there's no way to convert a metre of A, B, or C, into any
fraction of a tonne. Considering that a metre of A costs $800,
it better have a weight of a considerable fraction of a tonne,
or you can't make a profit.

> i just want help with the starting variables:
> i've started to write:: (x1, x2 representing the 2 lines)
> z=790[x1]+700[x2]-800(0.5[x1]+0.2[x2])-600(0.2[x1]+0.2[x2])-400(0.3[x1]+0.3[x2]
>
> subject to
> amount of A : 0.5[x1] + 0.2[x2] < 100
> amount of B : 0.2[x1] + 0.5[x2] < 100
> amount of C : 0.3[x1] + 0.3[x2] < 75
>
> can anyone tell me if i'm on the right track? i don't know where to put the purchasing prices!!!

And I think that this impacts your setup here. If a metre of A and a
metre of B
do not mass the same, then your formula for z is not correct. Further,
if a metre
total of inputs does not convert to a tonne of outputs, it's not
correct.

So, basically, if you are going to type in your homework, you should
type in all
of it. For example, what's your instructor's email so we can ask
clarifying
questions directly?
Socks

.

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