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.

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) (Show Source): You can put this solution on YOUR website!

Solve for y to get

Then we plug it into the expression to get

=
=

The goal is to make 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 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)

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