Question

There are 2 types: 150 kg of Nut and 90 kg of Razel
Mix 1: (1/2) N + (1/2) R, price $7/ 1kg

Mix 2: (3/4) N + (1/4) R, price $9 / kg
How many of Mix 1 and Mix 2 makes the total price maximum?
Please, help

Answer 1

what does this mean "How many of Mix 1 and Mix 2 "

Answer 2

This is what I try
Let X be number of Mix 1, then number of Nut in Mix 1 is 1/2 X
Let Y be number of Mix 2, then number of Nut in Mix 2 is 3/4 Y
then (1/2 )X+ (3/4)Y = 150, right?

Answer 3

Same, we have number of Razel in Mix 1 is 1/2 X
number of Razel in Mix 2 is 1/4 Y
and 1/2 X + 1/4 Y =70

Answer 4

But I don't know how to argue for 7x + 9y max. It is not calculus. It is algebra, hence we can't use derivative to find max of a function

Answer 5

is this linear programming question

Answer 6

Yes. it is

Answer 7

Well I know how I would do this but I don't know if it's "linear programming" but it is linear algebra and doesn't use calculus.

I'd diagonalize the matrix so that I could find out individually how much each nut was $/kg and then the nut that's highest is the one that maximizes it.

Answer 8

Please, show me how to solve it. May be I can derive from it.

Answer 9

Sure, there are really two ways to go about doing this, maybe diagonalization is more difficult I think since it requires you to find eigenvalues and eigenvectors which we can do without. Instead we can just solve the system of equations since there's only two.

so:

1) Solve the system of equations for N and R.

2) whichever is larger, N or R is what maximizes the sum, by only picking 100% of one and 0% of the other since one is always larger in price than the other.

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So hypothetically speaking if you have A = 90 and B=40 then what coefficients maximize this sum:

\[xA+yB=?\]

with the constraint that \(x+y=1\) (which is what we have for our total amount)

Well since A>B we know x=1 and y=0 maximizes the sum.
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I'm not really sure how else to explain this since I don't really know what you have issues with or what you're really expected to be doing or what.

Answer 10

I don't think it is the answer. As you said, x=1 y =0 , that is Mix 1 take over all of the nut and razel, but then 70 kg Razel need 70 kg nut to get 140 bags of Mix 1, how about 80 kg Nut?
Moreover, 140 bags*7= 980 is maximum value we can get if we let y=0
It is easy to find another solution > 980. Let say :
If I take just 50 kg of Nut and 50 kg of Razel to get 100 bags of Mix 1 *$7 =700
Amount of Razel left is 20kg, I can make 50 more bags of Mix2
and 50 more bags of Mix 2 needs 37.5kg of Nut accompanied with 20kg Razel to form 50 bags of Mix 2 * 9 = 450
Now, I have total $1150 > 980
Hence x=1 y =0 is not the best answer.

Answer 11

Notice: we have to use all of the Nut and Razel with an appropriate ratio to get the maximum of price.

Answer 12

Maximize
\(7x + 9y\)

subject to :
\(x\ge 0\)
\(y\ge 0\)
\(1/2x + 3/4y \le 150 \)
\(1/2x + 1/4y \le 90 \)