<s>[INST] 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!
 [/INST] 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 $k = \boxed{\sqrt{29}}$.
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