Let $x$ and $y$ be real numbers such that $x^2 + y^2 = 1$. Find the maximum value of $2x - 5y$. Let $x$ and $y$ be real numbers such that $x^2 + y^2 = 1$. Find the maximum value of $2x - 5y$. 
I do know how to solve this problem using trigonometry, however I need to solve it by using vectors. Except for defining vector $v=(x, y)$, such that $||v||=1$, I do not see any way in which I could apply vectors to solving this problem. Any ideas are appreciated. Thank you!
 A: Let $\vec a=(2,-5), \vec b=(x,y)$.
$2x-5y=\vec a\cdot \vec b=\sqrt{2^2+(-5)^2}\cdot\sqrt{x^2+y^2}\cos\theta\le\sqrt{29}$ where $\theta$ is the angle between the two vectors.
A: By the Cauchy-Schwarz inequality,
$$ \left|2x-5y\right| \leq \sqrt{2^2+5^2}\cdot\sqrt{x^2+y^2} = \sqrt{29} $$
and equality is attained when $(x,y)=\lambda\cdot(2,-5)$.
A: $\bf{My\; Solution::}$ Using Coordinate Geometry.
Let $2x-5y=k\;,$ Now value of $k$ is $\bf{Max.}$ when line is Tangent to the Circle $x^2+y^2=1$
Whose center is at $(0,0)$ and Radius $ = 1.$
So $\bf{Distance \; of \; line\; from \; (0,0)} = $ Radius of Circle.
So $\displaystyle \left|\frac{2(0)-5(0)-k}{\sqrt{2^2+(-5)^2}}\right| = 1\Rightarrow k=\pm \sqrt{29}$
So $\bf{(2x-3y)_{Max.}} = \sqrt{29}.$
A: Hint: consider the function
$$f(x,y,\lambda)=2x-5y+\lambda(x^2+y^2-1)$$
and solve the system
$$f_x=0$$
$$f_y=0$$
$$f_\lambda=0$$
A: The maximum value $k$ occurs when the line $2x-5y = k$ is tangent to the circle $x^2+y^2=1$.
Solving for $x$, we get $x = \frac{1}{2} (k+5y)$, so $\frac{1}{4}(k^2+10ky+25y^2) + y^2 = 1$. Now set up the quadratic and set $\Delta = 0$, as we only want there to be one value of $y$ for the intersection of the circle and line:
$$k^2+10ky + 25y^2 + 4y^2 = 4$$
$$\Rightarrow 29y^2+10ky+(k^2-4)=0$$
$$\Delta = b^2-4ac = 0: (10k)^2-4(29)(k^2-4)=0$$
$$\Rightarrow -16k^2+464=0, k = ±\sqrt{\frac{464}{16}} = ±\sqrt{29}$$
and we are looking for the maximum value of $k$. Thus $k = \boxed{\sqrt{29}}$.
