# Thread: min/max for partial differential

1. ## min/max for partial differential

$U = x + 2xy + y$

What's the minimum or maximum of U?

Differentiating I get:

$dU = (1 + 2y)dx + (2x + 1)dy$

So, is it correct to say that both the partial differentials are constant, therefore there is no minimum or maximum?

-Scott

2. Hello, Scott!

$U(x,y) \:= \:x + 2xy + y$

What's the minimum or maximum of $U$ ?
We're expected to equate the partial derivaitves to zero, and solve.

. . $U_x \:=\:1 + 2y \:=\:0\quad\Rightarrow\quad y \:=\:-\frac{1}{2}$

. . $U_y \:=\:2x+1\:=\:0\quad\Rightarrow\quad x \:=\:-\frac{1}{2}$

$U\left(\text{-}\frac{1}{2},\,\text{-}\frac{1}{2}\right) \:=\:-\frac{1}{2} + 2\left(\text{-}\frac{1}{2}\right)\left(\text{-}\frac{1}{2}\right) - \frac{1}{2} \:=\:-\frac{1}{2}$

Since I could see that $U(x,y)$ can increase without bound,
. . I suspected that we had found a minimum value.

Then I tried the Second Partials Test . . .

. . . . $U_{xx} \:=\:0\qquad U_{yy} \:=\:0 \qquad U_{xy} \:=\:U_{yx} \:=\:2$

$D \;=\;\left(U_{xx}\right)\left(U_{yy}\right) - \left(U_{xy}\right)^2 \;=\;0\cdot0 - 2^2\;=\;-4$ . . . . negative

Therefore: . $\left(-\frac{1}{2},\:-\frac{1}{2},\:-\frac{1}{2}\right)$ .is a saddle point.

. . The function has neither a maximum nor a minimum.