Maximizing u(x, y) with A and B constraints: Tips from Gekkoo

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To maximize the function u(x, y) = x^α * y^β under the constraint Ax + By = m, the method of Lagrange multipliers is recommended as an effective approach. Alternatively, one can express y in terms of x using the constraint, leading to a single-variable function. The next steps involve differentiating this function with respect to x and setting the first-order condition (FOC) to zero to find critical points. The user is struggling with deriving the FOC correctly and seeks assistance in solving it. Understanding these steps is crucial for identifying candidate solutions and checking the second-order conditions for maxima.
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Homework Statement



I need help to maximize the below function:

Homework Equations



Maximize u(x, y) = x^α * y^β subject to Ax + By = m

Any help is greatly appreciated!

/ Gekkoo
 
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I'm a big fan of "Lagrange Multipliers" but if you don't know that method, you could just write y= (m- Ax)/B so that x^\alpha*y^\beta= x^\alpha*(m-Ax)^\beta/B^\beta. Now, do you know how to find maxima and minima for that?
 
Thanks for your answers.

1 Solve constraint for y:

y=(m-Ax)/B

2 Plug into objective function:

u=x^α*[(m-Ax)/B]^β

3 Diff w.r.t. x & equate to zero to get critical point:

FOC: x^α*ln(x)*?=0

4 Solve FOC for x:

x=?

5 Plug that into constraint to get value for y:

y = (m-A[?])/B

6 Than I have a candidate solution & need to check SOC of objective function w.r.t x!

But I fail to successfully derive FOC. Can anyone please help me out?
 
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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