Maximize and Minimize $\vec C$ s.t. $|\vec A + \vec B + \vec C| \le 2400$ Given three vectors: $|\vec A| = 2850$ and forms a $150^\circ$ angle with the positive $x$-axis, $|\vec B| = 650$ and forms a $60^\circ$ angle with the positive $x$-axis, $\vec C$ which has a positive $x$-component and zero $y$-component. I need to find the minimum and maximum values of $|\vec C|$ such that $|\vec A + \vec B + \vec C| \le 2400$.
I initially thought about using law of cosines,
\begin{align*} 
|\vec A + \vec B|^2 &= |\vec A|^2 + |\vec B|^2 - 2|\vec A||\vec B|\cos 90^\circ \\
|\vec A + \vec B|^2 &= 2850^2 + 650^2 - 0 \\
|\vec A + \vec B| &\approx 2923.183 \\
\end{align*}
And I could repeat this using $|\vec A + \vec B|$ and $|\vec C|$, to see what value of $|\vec C|$ gives me $2400$, but I don't see how this gives me the minimum and maximum possible values.
What approach should I be taking to find the min and max values?
 A: All three vectors are essentially given, so maybe we can simply write them all in components, add them together, and solve the given inequality for the unknown component of $\vec C$?
Here's what I mean. From its description,
$$\vec A = \left\langle 2850\cos(150^{\circ}),2850\sin(150^{\circ}) \right\rangle = \left\langle -1425\sqrt{3},1425 \right\rangle.$$
Similarly, you can find vector $\vec B$. And $\vec C=\langle x,0 \rangle$ with the unknown component $x$. Add them together to form $\vec A+\vec B+\vec C$, and then set up the inequality $\left|\vec A+\vec B+\vec C\right|\le2400$ to solve for the unknown $x$. By the way, to avoid the square root in the norm of a vector formula, you should square both sides to be solving $\left|\vec A+\vec B+\vec C\right|^2\le5760000$.
A: Hint:
$$\vec A = 2850\cos(150^\circ) \hat{i} + 2850\sin(150^\circ) \hat{j}$$
$$\vec B = 650\cos(60^\circ) \hat{i} + 650\sin(60^\circ) \hat{j}$$
$$\vec C = c \hat{i}$$
Formulate $|\vec A + \vec B + \vec C|\le 2400$ as $\alpha c^2+\beta c+\delta \le 0$ to compute the minimum and maximum values of $c=|\vec C|$.
