Precalculus Vectors

Question

Let $x$ and $y$ be real numbers such that $x^2 + y^2 = 1$. Find the maximum value of $2x - 5y$.

I don't understand how to incorporate vectors into solving this problem. Or for that matter, how to solve the problem.

Answer 1

Here's a way to do it, but not with vectors. Let $x=\cos\theta,y=\sin\theta$ for any real $\theta$. This satisfies $x^2+y^2=1$, and does not change the possible values of $x$ and $y$. So now the goal is to maximise $2x-5y$. We can do this as such:

$$2x-5y=2\cos\theta-5\sin\theta=\sqrt{29}\left(\frac2{\sqrt{29}}\cos\theta-\frac5{\sqrt{29}}\sin\theta\right)$$

Note that if $\sin x=\frac2{\sqrt{29}},\cos x=\frac5{\sqrt{29}}$, then $\tan x=\frac25$. Let $a=\arctan\frac25$, then

$$\sqrt{29}\left(\frac2{\sqrt{29}}\cos\theta-\frac5{\sqrt{29}}\sin\theta\right)=\sqrt{29}\left(\sin a\cos\theta-\cos a\sin\theta\right)=\sqrt{29}\sin(a-\theta)$$

Which is maximised when $\sin(a-\theta)=1$. Hence the maximum value of $2x-5y$ is $\sqrt{29}$.