Formulate this as a linear programming problem

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Homework Statement
See attached: this decision maths text book I realized has a number of wrong notations and solutions in various topics:critical path etc

Just counter check the phrase "twice as many pockets as desktop"

Is that correct? Is it not firstly, ##2y=x##
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Decision maths
See attached ...
 

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I believe ##x## is used for desktop diaries and ##y## is used for pocket diaries.

They key word is “at least”

“They will need at least twice as many pocket diaries as desktop diaries”……translation: “the number of pocket diaries must be greater than or equal to 2 times the number of desktop diaries”

So

##y \geq 2x##
 
And this one, sorry, am using phone to type...
Number 2,
For kruskal algorithm, CE shouldn't be connected as it forms a cycle, correct? ...or a cycle refers to connected vertices? am I missing something. See my sketch...
 

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Last edited:
chwala said:
Just counter check the phrase "twice as many pockets as desktop"

Is that correct? Is it not firstly, ##2y=x##
The phrase “there is twice as many ## y ## as ## x ##” means ## y=2x ##.
 
chwala said:
...or a cycle refers to connected vertices?
Yes, it does.

The edge CE is included into the minimum spanning tree too.
 
Gavran said:
Yes, it does.

The edge CE is included into the minimum spanning tree too.
I thought CE shouldn't be connected as shown on text, as that does not conform to the kruskal algorithm. I stand to be corrected.

Just seen that intersection can occur as long as the vertices are not connected to form loop, in that case the textbook is correct. Cheers man.
 
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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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