Question

Given the function:

f(x1,x2) = 2/(x1x2) + x1x2 + x1,

Show that there is no solution of minimizing this function over x1 > 0, x2 > 0. Can f(x1,x2) ever be smaller than 2sqrt2?

Answer 1

Answer on question 35970 – Math – Multivariable calculus

Given the function $g(t1, t2) = 2 / (t1t2) + t1t2 + t1$. Show that there is no solution to the problem of minimizing this function over $t1 > 0$, $t2 > 0$. Can $g(t1, t2)$ ever be smaller than $2\sqrt{2}$?

Solution

$$\left\{ \begin{array}{c} g (t _ {1}, t _ {2}) = \frac {2}{t _ {1} t _ {2}} + t _ {1} t _ {2} + t _ {1} \to \min \\ t _ {1} > 0 \\ t _ {2} > 0 \end{array} \right.$$

According to the necessary condition for an extrema of function of two variables we need to find the solution of the following system

$$\left\{ \begin{array}{l} \frac {\partial g \left(t _ {1} , t _ {2}\right)}{\partial t _ {1}} = 0 \\ \frac {\partial g \left(t _ {1} , t _ {2}\right)}{\partial t _ {2}} = 0 \end{array} \right. \quad \left\{ \begin{array}{l} - \frac {2}{t _ {1} ^ {2} t _ {2}} + t _ {2} + 1 = 0 \\ - \frac {2}{t _ {1} t _ {2} ^ {2}} + t _ {1} = 0 \end{array} \right.$$

From the second equation we get

$$t _ {1} ^ {2} = \frac {2}{t _ {2} ^ {2}}$$

Substitute this into the first equation

$$- \frac {2 t _ {2} ^ {2}}{2 t _ {2}} + t _ {2} + 1 = 0, \qquad 1 = 0.$$

This equation has no solutions.

Therefore, the system of equations has no solution and the function $g(t_{1},t_{2})$ has no extrema.

Now let us consider the function $f(t_{1},t_{2}) = \frac{2}{t_{1}t_{2}} + t_{1}t_{2}$, where $t_1 > 0, t_2 > 0$. Obviously that $f(t_{1},t_{2}) < g(t_{1},t_{2})$.

Find the minimum of the function $f(t_{1},t_{2}) = \frac{2}{t_{1}t_{2}} + t_{1}t_{2}$. Let $t_1t_2 = x$ then $f(x) = \frac{2}{x} +x$.

$$f ^ {\prime} (x) = - \frac {2}{x ^ {2}} + 1 = \frac {x ^ {2} - 2}{x ^ {2}} = 0$$ $$x = \sqrt {2}.$$

At the point $\sqrt{2}$ this function has a minimum.

$$f \left(\sqrt {2}\right) = \frac {2}{\sqrt {2}} + \sqrt {2} = \frac {4}{\sqrt {2}} = 2 \sqrt {2}.$$

And we get $2\sqrt{2} \leq f(t_{1}, t_{2}) < g(t_{1}, t_{2})$.

**Answer**: the function $g(t_{1}, t_{2})$ has no extrema and never be smaller than $2\sqrt{2}$.