Question 1209306
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Solve {{{xy = 4/3}}} for y to get {{{y = 4/(3x)}}}


Then we plug it into the expression to get
{{{2x + 6y + x^2 + 3y^2}}}
= {{{2x + 6*(4/(3x)) + x^2 + 3*(4/(3x))^2}}}
= {{{(3x^4+6x^3+24x+16)/(3x^2)}}}


The goal is to make {{{f(x) = (3x^4+6x^3+24x+16)/(3x^2)}}} 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 <a href="https://www.desmos.com/calculator">Desmos</a> and <a href="https://www.geogebra.org/calculator">GeoGebra</a>
There are many others to pick from.


Here's the Desmos graph
<a href="https://www.desmos.com/calculator/nieyc6olhx">https://www.desmos.com/calculator/nieyc6olhx</a>
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(x)}}} 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: <font color=red>12.82996 (approximate)</font>
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