Quadratic equation solving with constraint.

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Homework Help Overview

The problem involves solving a quadratic equation with a constraint related to optimization. The equation is given as -0.03x^2 + 40x - 0.02y^2 + 5y + 55000 = 0, with the constraint x + y = 3000.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss substituting the constraint into the quadratic equation and explore the geometric interpretation of the equation and constraint.

Discussion Status

Some participants have provided guidance on how to approach the substitution of the constraint into the quadratic equation. There is acknowledgment of the geometric relationship between the quadratic and the constraint.

Contextual Notes

The original poster expresses uncertainty about the final steps needed to solve the problem, despite having worked through significant prior calculations. There is a mention of previous optimization methods used, such as Lagrangians and Cramer's rule.

qwerty11
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Homework Statement


So I've worked this problem for awhile. Its a several page of math problem with optimization, legrangians, cramers rule, etc to get to this point. All I need to do now is by hand solve this equation for x and y with the constraint.


Homework Equations



-0.03x^2 + 40x - 0.02y^2 + 5y+ 55000 = 0
Subject to the constraint: x+y=3000


The Attempt at a Solution



I could put up the pages of work I've done to get to this point, but it really doesn't have any relevant info to help get to this point. I have the answers which is x=1100 & y=1900. Just don't know how to do this last step I've got to.

Thanks so much for the help!
 
Last edited by a moderator:
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qwerty11 said:
-0.03x^2 + 40x - 0.02y^2 + 5y+ 55000 = 0
Subject to the constraint: x+y=3000

If your constraint is x+y=3000 then you just need to plug y = 3000-x into the quadratic.
 
Last edited by a moderator:
Geometrically, the quadratic equation in x and y represents an ellipse, and the constraint represents a straight line, so you will either get no solutions, one solution, or two solutions.
 
Mentallic said:
If your constraint is x+y=3000 then you just need to plug y = 3000-x into the quadratic.

Thanks so much! This is exactly what was messing me up. Dono why I didn't think of that.

Thanks again!
 
qwerty11 said:
Thanks so much! This is exactly what was messing me up. Dono why I didn't think of that.

Thanks again!

You're welcome :smile:
 

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