Question

Find the maximum value of s = xy + yz + xz where x+y+z=9.​

Answer 1

From the constraint, we have

$x+y+z=9 \implies z = 9-x-y$

so that $s$ depends only on $x,y$.

$s = g(x,y) = xy + y(9-x-y) + x(9-x-y) = 9y - y^2 + 9x - x^2 - xy$

Find the critical points of $g$.

$\dfrac{\partial g}{\partial x} = 9 - 2x - y = 0 \implies 2x + y = 9$

$\dfrac{\partial g}{\partial y} = 9 - 2y - x = 0$

Using the given constraint again, we have the condition

$x+y+z = 2x+y \implies x=z$

so that

$x = 9 - x - y \implies y = 9 - 2x$

and $s$ depends only on $x$.

$s = h(x) = 9(9-2x) - (9-2x)^2 + 9x - x^2 - x(9-2x) = 18x - 3x^2$

Find the critical points of $h$.

$\dfrac{dh}{dx} = 18 - 6x = 0 \implies x=3$

It follows that $y = 9-2\cdot3 = 3$ and $z=3$, so the only critical point of $s$ is at (3, 3, 3).

Differentiate $h$ again and check the sign of the second derivative at the critical point.

$\dfrac{d^2h}{dx^2} = -6 < 0$

for all $x$, which indicates a maximum.

We find that

$\max\left\{xy+yz+xz \mid x+y+z=9\right\} = \boxed{27} \text{ at } (x,y,z) = (3,3,3)$