Solving Math Formulaes: X/5 to X-5 with Lagranges Equation

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SUMMARY

The discussion centers on transforming the expression X/5 into X-5 and applying Lagrange's equation to solve a mathematical problem. The user initially derives the equation 5y = 3000 - x, leading to y = 600 - x/5. They then formulate the function f(x,y) = 100x^1/2 + 100y^1/2 - 1000, which simplifies to 100x^1/2 + 100(600 - x/5)^1/2 - 1000. The consensus is that to solve this, one must isolate the square roots and square both sides of the equation, potentially requiring multiple squarings.

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Homework Statement



Does anybody know of some good mathematics formula books?

i am currently stuck with: X/5

Can i turn this into something else, like: X-5?

i got this from the equation: X+ 5y = 3000

then: 5y = 3000 - x

y = 600 - x/5

Now i have to put this into a Lagranges equation:

f(x,y) = 100x^1/2 + 100y^1/2-1000

which i now get: 100x^1/2 + 100 (600-x/5)^1/2-1000

how do i solve this?


Homework Equations





The Attempt at a Solution

 
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What, exactly are you trying to do? You talk about "solving" an equation, but you have no equation. IF what you have is set equal to something, typically, what you would need to do is isolate the square roots on opposite sides of the equals sign and square both sides. You probably will need to square twice.
 
What you're asking for doesn't exist.
storoi1990 said:
Does anybody know of some good mathematics formula books?
i am currently stuck with: X/5
Can i turn this into something else, like: X-5?
About the only thing you can do with x/5 is to write it as a product: x * (1/5). About the only thing you can do with x - 5 is to write it as a sum: x + (-5). You can always write a quotient of two expressions as a product of one of them times the reciprocal of the other, and you can always write a sum of two expressions as one of them plus the additive inverse of the other, but that's about it.
 

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