Double maxed out

Using Lagrange multipliers, we need to solve the equation $\left(\begin{array}{c} 2x \\ z^3 \\ 3yz^2 \end{array}\right) - \lambda\left(\begin{array}{c} 2x \\ 2y \\ 2z \end{array} \right) \; = \; \left(\begin{array}{c}0\\0\\0 \end{array}\right)$ and so $2x(1-\lambda) \; = \; z^3 - 2\lambda y \; = \; z(3yz - 2\lambda) \; = \; 0$

1. If $z=0$ , then $\lambda y = x(1-\lambda) = 0$ . If $\lambda = 0$ then $x=0$ and $y = r^{\frac14}$ , and the matching extremal value of $F$ is $0$ . If $\lambda \neq 0$ then $y=0$ , so that $x = r^{\frac14}$ and $\lambda=1$ , and the matching extremal value of $F$ is $r^{\frac12}$ .

2. If $z \neq 0$ then $\lambda = \tfrac32yz$ and $x(2 - 3yz) = z(z^2 - 3y^2) = 0$ . Thus $z^2 = 3y^2$ , so $z = y\sqrt{3}$ , so $x(2 - 3\sqrt{3}y^2) = 0$ . If $x=0$ then $y = \tfrac12r^{\frac14}$ and $z = \tfrac{\sqrt{3}}{2}r^{\frac14}$ , and the matching extremal value of $F$ is $\tfrac{3\sqrt{3}}{16}r$ . If $x \neq 0$ then $y \; = \; \sqrt{\frac{2}{3\sqrt{3}}} \qquad z \; = \; \sqrt{\frac{2}{\sqrt{3}}}$ and hence $x^2 \; = \; r^{\frac12} - \frac{8}{3\sqrt{3}}$ The matching extremal value of $F$ is $r^{\frac12} - \frac{4}{3\sqrt{3}}$ which is smaller than $r^{\frac12}$ .

Thus $r^\star$ is identified by the requirement that $(r^\star)^{\frac12} \,=\, \frac{3\sqrt{3}}{16}r^\star$ , which implies that $r^\star \,=\, \frac{256}{27}$ . The required answer is $256+27 = 283$ .

First let $R=y^2+z^2=\sqrt r -x^2$ , which ranges from $0$ to $\sqrt r$ . We'll find the largest value of $F(x,y,z)=x^2+yz^3 =\sqrt r -R +yz^3$ for a fixed $R>0$ . The key here is to maximize the function $g(\theta) = \cos \theta \sin^3 \theta$ with domain $[0,\pi/2]$ . We note that $g'( \theta)$ is

$3 \cos^2 \theta \sin ^2 \theta - \sin^4 \theta =\sin ^2 \theta (3 \cos^2 \theta - \sin^2 \theta) =\sin ^2 \theta (4 \cos^2 \theta - 1)$ .

Also note that at the endpoints, $g(0)=0=g(\pi/2)$ , and the only other critical point is $\theta=\pi/3$ . Hence $g$ has a unique maximum, $g(\pi/3)= \left ( \frac 1 2 \right ) \left ( \frac {\sqrt{3}} 2 \right ) ^{3} =\frac {3\sqrt{3}} {16}$

From the above result it follows that, for a fixed $R\ge0$ , the largest value of $F(x,y,z)=x^2+yz^3 =\sqrt r -R +yz^3$ is:

$m(R)=\sqrt r -R + \frac {3R^2\sqrt{3}} {16}$ .

(Note that this is trivial for $R=0$ .) Since $m(R)$ is an upward-concave parabola (as a function of $R$ ) it achieves its maximum at one or both of the endpoints $R=0$ , $R= \sqrt r$ . Therefore $r=r^ *$ if and only if $m(0)=m(\sqrt {r})$ . Squaring both sides, this equation becomes $r= \left (\frac {3r\sqrt{3}} {16}\right) ^2$ and since $r\ne 0$ this means that $r^*= \frac {256} {27}$ .

283

283