Is (3/4)*(a^2/c) less than a with multiple inequalities?

AI Thread Summary
The discussion revolves around the inequalities involving two variables, x and y, and two positive constants, a and c. The main question is whether the expression (3/4)*(a^2/c) is less than a, given the inequalities x + 2y ≤ (3/4)*(a^2/c) and x + 2y < a. Participants highlight the need for a complete problem statement to clarify the relationships between the variables and constants. The conversation also touches on the implications of inequalities, questioning if certain conditions lead to contradictory results. Overall, the discussion emphasizes the importance of clear definitions and logical reasoning in solving inequalities.
monsmatglad
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Homework Statement


I for some reason can't seem do become sure of this.
There are 2 variables x and y. And two constants, a and c, which are both positive.

Homework Equations


x+2y ≤ (3/4)*(a^2/c)
x + 2y < a

The Attempt at a Solution


Does this mean that: (3/4)*(a^2/c) < a ?

Mons
 
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monsmatglad said:

Homework Statement


I for some reason can't seem do become sure of this.
There are 2 variables x and y. And two constants, a and c, which are both positive.

Homework Equations


x+2y ≤ (3/4)*(a^2/c)
x + 2y < a

The Attempt at a Solution


Does this mean that: (3/4)*(a^2/c) < a ?

Mons
Please give a complete statement of the problem which you're trying to solve.
 
monsmatglad said:

Homework Statement


I for some reason can't seem do become sure of this.
There are 2 variables x and y. And two constants, a and c, which are both positive.

Homework Equations


x+2y ≤ (3/4)*(a^2/c)
x + 2y < a

The Attempt at a Solution


Does this mean that: (3/4)*(a^2/c) < a ?

Mons
Do the inequalities ##A \leq 200## and ##A < 100## imply ##200 < 100?##
 
@monsmatglad, please check your Inbox. I explained why the inequality isn't necessarily true, based on the info you provided in your earlier thread.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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