Maximizing in a Real Quadratic

We have

$2b+3c=195-a$ and $b^{2} + c^{2} =2925-a^{2}$

By using Cauchy - Schwartz Inequality, we obtain

$(b^{2}+c^{2})(4+9)\ge(2b+3c)$

Now,

$(2925-a^{2})(4+9)\ge(195-a)$

Hence,

$14a^{2} - 2(195)\ge0$

Solving for $a$ gives

$a\le0$ or $a\ge\frac{195}{7}$

Which means that $p+q=195+7=202$

We rearrange the equations to get $2b+3c=195-a$ and $b^2+c^2=2925-a^2$ . By Cauchy-Schwarz Inequality, $(4+9)(b^2+c^2) \geq (2b+3c)^2$ which using the above substitutions give us $13(2925-a^2) \geq (195-a)^2$

$38025 - 13a^2 \geq a^2 - 390a + 38025$

$14a^2 - 390a \leq 0$

$a \leq \frac{195}{7}$ since $a$ is positive with equality when $b=\frac{180}{7}$ and $c = \frac{270}{7}$ based on the equality condition of Cauchy-Schwarz that $\frac{4}{9}=\frac{b^2}{c^2}$

Hence the answer is $195+7=202$

We want to maximize $a$ , subject to two constraints $a+2b+3c=195$ and $a^2+b^2+c^2=2925$ . We can use Lagrange multipliers.

Let $f(a,b,c,\lambda,\mu)=a-\lambda(a+2b+3c-195)-\mu(a^2+b^2+c^2-2925)$ . To get the optimal value, we should equate all partial differentials to zero.

$f_a=1-\lambda-2a\mu=0$

$f_b=-2\lambda-2b\mu=0$

$f_c=-3\lambda-2c\mu=0$

$f_{\lambda}=-a-2b-3c+195=0$

$f_{\mu}=-a^2-b^2-c^2+2925=0$

Given these, we can solve the following using $f_a=0$ , $f_b=0$ , and $f_c=0$ :

$a=\frac{\lambda-1}{2\mu}$

$b=\frac{2\lambda}{2\mu}$

$c=\frac{3\lambda}{2\mu}$

Thus, using $f_{\lambda}$ ,

$\frac{\lambda-1}{2\mu}+2\left(\frac{2\lambda}{2\mu}\right)+3\left(\frac{3\lambda}{2\mu}\right)=195$

$\frac{\lambda-1+4\lambda+9\lambda}{2\mu}=195$

$\mu=\frac{14\lambda-1}{390}$

Using $f_{\mu}=0$ ,

$\frac{195^2}{(14\lambda-1)^2} \left((\lambda-1)^2+(2\lambda)^2+(3\lambda)^2 \right)=2925$

Solving this, we will get $\lambda=1$ or $\lambda=-^6/_7$ .

If $\lambda=1$ , $a=0$ , while if $\lambda=-{}^6/_7$ , $a={}^{195}/_7$ .

Thus, the maximum value of $a$ is ${}^{195}/_7$ . Hence, $p+q=195+7=202$ .

Note that $(b^2+c^2)(4+9)\ge(2b+3c)^2$ by Cauchy-Schwarz. This means $(2925-a^2)(13)\ge (195-a)^2$ , or $2a(195-7a)\ge 0$ . This is turn gives $0\le a\le \frac{195}{7}$ , so the answer is $195+7=\boxed{202}$ . Note that equality is achievable when $a=\frac{195}{7}, b=\frac{180}{7}, c=\frac{270}{7}$ .

We have b= $\frac{195-a-3c}{2}$ and $b^2+c^2=2925-a^2$ . The above two equations give us $13c^2-6c(195-a)+(195-a)^2+4a^2-11700$ = 0 The discriminant of the above equation becomes $6240a-224a^2$ = $32(195a-7a^2)$ which must be graeter than or equal to 0. So $a \leq \frac{195}{7}$ which means that the solution to the question is 202.

We have $2b+3c=195-a$ and $b^2+c^2=2925-a^2$ . By Cauchy Schwarz Inequality, we have $(4+9)(b^2+c^2)\geq (2|b|+3|c|)^2\geq (2b+3c)^2$ , which gives $13(2925-a^2)\geq (195-a)^2$ or $38025-13a^2\geq 38025-390a+a^2$ . Thus $14a^2-390a\leq 0$ , which gives $2a(7a-195)\leq 0$ , so $0\leq a\leq \frac{195}{7}$ . Therefore, the maximum value of $a$ is $\frac{195}{7}$ , and $p+q=195+7=202$ .

(195-a)^2<=13(2925-a^2)