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TL DR
Not at all confident in his reflexive opinions of thingsRegistered User regular

You guys helped me ace my math test last time around, now I must call upon you again.

I'm constructing a table to algebraically solve this problem:

Maximize P = 3x + 4y + 2z

Subject to x + y + z <= 10

x + 2y +2z <= 14

I rewrite with slack variables S and T, and all is going well, until I have to find the values for Y and Z when X, S, and T are zero variables.

So my problem would look like this:

y + z = 10

2y + 2z = 14

And I am stuck.

All help is appreciated, thanks in advance.

I'm constructing a table to algebraically solve this problem:

Maximize P = 3x + 4y + 2z

Subject to x + y + z <= 10

x + 2y +2z <= 14

I rewrite with slack variables S and T, and all is going well, until I have to find the values for Y and Z when X, S, and T are zero variables.

So my problem would look like this:

y + z = 10

2y + 2z = 14

And I am stuck.

All help is appreciated, thanks in advance.

0

## Posts

So, y = 10 - z

Plug in 2(10-z) + 2 z = 14

Solve accordingly?

Edit: I realized that my math yields a zero. I think you must've mixed something up earlier on, because y + z = 10 and y+z = 7

AphostileonNothing. Matters.x + y + z + s = 10

x + 2y+ 2z+ t = 14

And when x, s, and t all = 0, we're left with

0 + y + z + 0 = 10

0 + 2y+ 2z+ 0 = 14

So taking you're suggestion and discarding the zero variables, I'd get something like

y = 10 - z

which, plugged into the 2nd equation, results in

2 (10 - z) +2z = 14

or

20 = 14

O_o:|

TL DRonThis one is asking me to minimize c = 16x + 42y + 5z

subject to 2x + 6y >= 20

2x + 3y + z >= 40

by finding the dual.

Dual:

Maximize P= 20x + 40x

Subject to 2x + 2y <= 16

6x + 3y <= 42

y <= 5

Pivot, pivot, and I get

x = 9/2

y = 5

p = 290

Not hard. I just can't figure out the deal with that damn table in the other problem.

TL DRonx + y + z <= 10

x + 2y + 2z <= 14

Make the assumption that the solution will come from the boundaries (it will, cause there's only one critical point and yay calculus):

x + y + z = 10

x + 2y + 2z = 14

Or:

2x + 2y + 2z = 20

x + 2y + 2z = 14

Which leads to:

x = 6

So then y + z = 4 from either of our equations above.

Or: y = 4 - z

So: P = 18 + 16- 4z + 2z

= 34 - 2z

Making the assumption that the variables cannot be nonnegative, z then = 0, y = 4 and P's maximum value is 34.

Alternately:

It's clear that x <= 6 and y + z <= 4, right? Because otherwise these equations wouldn't be used to their fullest potential and that's no good at all.

So you've got: P = 3x + 4y + 2z

Group it instead as follows:

P = 3x + 2y + 2(y + z)

The first term is maximized if x = 6, the third term is maximized when y + z = 4, and the second term is maximized when y is the biggest it can be such that y + z <= 4, i.e. y = 4.

And again we get x = 6, y =4, z = 0 and P = 34 as the maximum. However, again we assume the answer only has non-negative values for x, y, and z.

Hope that helps.

enlightenedbumonAcutally you are correct and it's not an assumption since the question asks you to maximise P and if you use negative values it will bring down the value of P making the solution invalid.

Blake TonSatans..... hints.....

It totally wouldn't! Check out: y = 100, z = - 96 and x = 6. It still fits all the constraints, but suddenly P is huge! namely, 18 + 400 - 192 = 226.

enlightenedbumonBlake TonSatans..... hints.....

FunkyWaltDoggon