Quadratic equation solving with constraint.

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To solve the quadratic equation -0.03x^2 + 40x - 0.02y^2 + 5y + 55000 = 0 with the constraint x + y = 3000, substitute y with 3000 - x. This substitution transforms the problem into a single-variable equation. The quadratic represents an ellipse, while the constraint is a straight line, leading to potentially zero, one, or two solutions. The final answers for x and y are x = 1100 and y = 1900. This method clarifies the solution process for the given constraint.
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!
 
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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.
 
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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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