Maximum of a Linear Function Subject to Linear-Quadratic Constraints

Let us simplify this optimization problem by taking the linear constraint $x + 2y - z = 10 \Rightarrow z = x + 2y - 10$ and substituting it into $f(x,y,z)$ to become $g(x,y) = 2x - 3y + 5(x + 2y - 10) = 7x + 7y -50$ and $h(x,y) = x^2 + y^2 + (x + 2y -10)^2 = 100.$

By taking Lagrange multipliers ( $\nabla g = \lambda \cdot \nabla h$ ), we now obtain:

$7 = \lambda \cdot [2x + 2(x + 2y - 10) \cdot 1];$

$7 = \lambda \cdot [2y + 2(x + 2y - 10) \cdot 2];$

or $x + (x + 2y -10) = y + 2(x + 2y - 10) \Rightarrow y = \frac{10}{3}.$

Now we solve for x in $h(x, \frac{10}{3}) = x^2 + (\frac{10}{3})^2 +[x +2(\frac{10}{3}) - 10]^2 = 100;$

or $x^2 + (\frac{10}{3})^2 + (x - \frac{10}{3})^2 = 100;$

or $2x^2 - \frac{20}{3}x + \frac{200}{9} = 100 \Rightarrow 18x^2 - 60x - 700 = 0 \Rightarrow x = \frac{60 \pm \sqrt{60^2 - 4(18)(-700)}}{36} \Rightarrow x = \frac{5}{3} \cdot (1 \pm \sqrt{15})$ .

Since we wish to maximize $g(x,y),$ we only accept the larger root $x = \frac{5}{3} \cdot (1 + \sqrt{15})$ . We now solve for $z = x + 2y -10 = \frac{5}{3} \cdot (1 + \sqrt{15}) + 2(\frac{10}{3}) - 10 = \frac{5}{3} \cdot (-1 + \sqrt{15})$ and for $f(x,y,z) = 2[\frac{5}{3} \cdot (1 + \sqrt{15}] - 3(\frac{10}{3}) + 5[ \frac{5}{3} \cdot (-1 + \sqrt{15})] = \frac{-45 + 35\sqrt{15}}{3}.$ Altogether, our final computation comes to:

$x + 2y + 3z + f_{max} = \frac{5}{3} \cdot (1 + \sqrt{15}) + 2(\frac{10}{3}) + 3[ \frac{5}{3} \cdot (-1 + \sqrt{15})] + \frac{-45 + 35\sqrt{15}}{3} = \boxed{\frac{55\sqrt{15} - 35}{3}}.$