From A Chinese Math Competition

We shall maximize $xy + kxz$ for $x,y,z > 0$ and $\sqrt{x^2+y^2} + z = 1$ , provided that $k > \sqrt{2}$ . If we write $x,y$ in polar coordinates, then $x \; = \; r\cos\theta \hspace{1cm} y \; = \; r\sin\theta \hspace{1cm} z \; = \; 1 - r \hspace{2cm} 0 < r < 1\;,\; 0 < \theta < \tfrac12\pi$ and we want to maximize $F(r,\theta) \; = \; r^2\sin\theta\cos\theta + kr(1-r)\cos\theta \hspace{2cm} 0 < r < 1 \;,\; 0 < \theta < \tfrac12\pi$ We find the maximum of $F$ by solving $\nabla F \; = \; \left(\begin{array}{c} 2r\sin\theta\cos\theta + k(1-2r)\cos\theta \\ r^2(\cos^2\theta - \sin^2\theta) - kr(1-r)\sin\theta \end{array}\right) \; = \; \mathbf{0}$ Looking at the first component, we deduce that $2r\sin\theta = k(2r-1)$ . Plugging this expression into the second component equation we obtain $\begin{aligned} r^2 - 2r^2\sin^2\theta - kr(1-r)\sin\theta & = \; 0 \\ r^2 - \tfrac12k^2(2r-1)^2 - \tfrac12k^2(2r-1)(1-r) & = \; 0 \\ \tfrac12r\big[2r - k^2(2r-1)\big] & = \; 0 \end{aligned}$ and hence we deduce that $r = \tfrac{k^2}{2(k^2-1)}$ . The condition that $k > \sqrt{2}$ ensures that this value of $r$ lies between $0$ and $1$ - if $0 < k \le \sqrt{2}$ there will be no maximum value of $F$ . Thus we deduce that $\sin\theta = \tfrac{1}{k}$ , and we obtain the maximum value of $F$ as $\frac{k^2}{4\sqrt{k^2-1}}$ For the particular case of this question, $k=4$ , and so the maximum value is $\frac{4}{\sqrt{15}}$ , making the answer $4+15= \boxed{19}$ .

We can get xy+4x(1-(x^2+y^2)^(1/2)). Let F(x,y)=xy+4x(1-(x^2+y^2)^(1/2)). The partial derivative of this function on X is "y+4-(4y^2+8x^2)/(y^2+x^2)^(1/2)".And the partial derivative of this function on Y is "(4xy)/(y^2+x^2)^(1/2)". Make them equal to 0.So we can get x=0，y=4/3 or x=2/(15)^(1/2),y=2/15 or x=-2/(15)^(1/2),y=2/15. Replace them back to the original function,and get the maximun is 4/(15)^(1/2).So A+B is 19. (Please don't try to find out the second derivative and use it,because it is too complicated.)