SOLUTION: Let x and y be nonnegative real numbers. If xy = 4/3, then find the minimum value of 2x + 6y + x^2 + 3y^2.

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Question 1209306: Let x and y be nonnegative real numbers. If xy = 4/3, then find the minimum value of 2x + 6y + x^2 + 3y^2.
Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

Solve xy+=+4%2F3 for y to get y+=+4%2F%283x%29

Then we plug it into the expression to get
2x+%2B+6y+%2B+x%5E2+%2B+3y%5E2
= 2x+%2B+6%2A%284%2F%283x%29%29+%2B+x%5E2+%2B+3%2A%284%2F%283x%29%29%5E2
= %283x%5E4%2B6x%5E3%2B24x%2B16%29%2F%283x%5E2%29

The goal is to make f%28x%29+=+%283x%5E4%2B6x%5E3%2B24x%2B16%29%2F%283x%5E2%29 as small as possible.
Note that x = 0 isn't allowed in the domain so we're focused on the interval x > 0.

You can use differential calculus or you can use a graphing calculator to determine the local minimum when x > 0

Two recommended graphing tools are Desmos and GeoGebra
There are many others to pick from.

Here's the Desmos graph
https://www.desmos.com/calculator/nieyc6olhx
When focusing on the interval x > 0 the lowest part of the red curve is approximately at (1.65132, 12.82996)
This is marked as the green point on the curve.

The y coordinate of this local minimum is the smallest value of f%28x%29 and also the smallest value of the original expression.

Side note: be sure not to mix up the y coordinate of the local minimum with the "y"s in either xy = 4/3 or 2x + 6y + x^2 + 3y^2. Those are different y values.



Answer: 12.82996 (approximate)