Is the Objective Function Always Nonnegative in Linear Optimization Problems?

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Homework Statement



Given the follow LOP P

I am just going to write down the obj function because that is most important for my question and the constraints aren't

w = y_1 - 2y_2 + y_3

I was asked to show that y = (2t, 3t,t)^t is a solution for all t\geq 0

So w = 2t - 2(3t) +t = 2t - 6t + t = -4t + t = -3t

Now initally I thought that as t \to \infty, -3t \to -\infty

I checked the key provided by my prof and he took t \to -\infty and -3t \to \infty

Is it because we always assume w >0?

The flaw I made is that I never consider t \to \pm \infty.




Homework Equations





The Attempt at a Solution

 
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In applied problems, the objective function is usually nonnegative, but in more theoretic presentations, I don't see why this needs to be true.
 


Mark44 said:
In applied problems, the objective function is usually nonnegative, but in more theoretic presentations, I don't see why this needs to be true.

I asked my prof today and he kinda said the same thing about "yes intuitively that is right, we want obj f > 0". Then he added a bunch of things that confused me even more...

He stated something like this

max z = -min(-z)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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