How Can You Minimize This Complex Fraction Given the Constraints?

[anemone]

Here is this week's POTW:

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Minimize \(\displaystyle \frac{2x^3+1}{4y(x-y)}\) given \(\displaystyle x\ge -\frac{1}{2}\) and \(\displaystyle \frac{x}{y}>1.\)

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[anemone]

Congratulations to the following members for their correct solution: (Smile)

1. Ackbach
2. Opalg
3. castor28
4. greg1313

Solution from Opalg:

Spoiler

Let $z = \dfrac xy$. Then $z>1$, and $y = \dfrac xz$. Therefore $$ \frac{2x^3+1}{4y(x-y)} = \frac{(2x^3+1)z^2}{4x(xz-x)} = \frac{2x^3+1}{4x^2}\cdot\frac{z^2}{z-1}.$$ The function $\frac{2x^3+1}{4x^2}$ is positive for all $x\geqslant -\frac12$, and the function $\frac{z^2}{z-1}$ is positive for all $z>1$. So to minimise the product of the two functions on those intervals, it is sufficient to minimise each function separately and then take the product.

The minimum of $\frac{2x^3+1}{4x^2}$ is $\frac34$ (occurring when $x = -\frac12$ and also when $x=1$). The minimum of $\frac{z^2}{z-1}$ is $4$ (occurring when $z=2$). So the minimum of their product is $3$, occurring at the points $(x,y) = \bigl(-\frac12,-\frac14\bigr)$ and $(x,y) = \bigl(1,\frac12\bigr)$.

Alternate solution from castor28:

Spoiler

Let us call the expression $f(x,y)$. The domain under consideration is delimited by the lines $y=0$, $y=x$, and $x=-\frac{1}{2}$. As the first two lines are not part of the domain, we can only have a minimum at an interior point or on the boundary $x=-\frac{1}{2}$.

We compute the partial derivative:

$$\displaystyle
\frac{\partial f}{\partial y} = -\frac{(2x^3+1)(x-2y)}{4y^2(x-y)^2}
$$
​

In the domain, this is $0$ only on the line $x=2y$; therefore, any interior minimum must be on that line.

We define:

$$\displaystyle
g(x) = f\left(x,\frac{x}{2}\right) = 2x + \frac{1}{x^2}
$$
​

This function has a single minimum at $x=1$ with value $g(1) = 3$.

We look now for a minimum on the boundary line $x=-\frac{1}{2}$. As this line is parallel to the $y$ axis, the condition $\partial f/\partial y=0$ still applies, and the only possible extremum is at $y=-\frac14$, where the value of the function is $3$. As $g'(-\frac12) =18>0$, this is indeed a minimum.

To summarize, we have two minima at $\left(1,\frac{1}{2}\right)$ and $\left(-\frac{1}{2},-\frac14\right)$, both with value $3$.