I have no idea how to do this at all.
"Find the point on the line $y=2x+3$ that is closest to the origin"
I am just not very smart or creative so I have no idea how to do this. I graphed it and I think I can assume that it will be closest to the origin near (1.5, 0) through (0, 3). Those are the x and y intercepts that I found (possibly correctly) and I think the formula for finding it close to the origin would be $x+y < 1.5$ since 1.5 is the closest point I have right now. Idealy I want x+y to equal 0 but it won't since that is probably not an intercept.
From here I have no idea what to do, I have tried a couple things but nothing even makes sense. I have never done a problem like this.
Best Answer
There are purely geometric ways to do the problem, but since it arose in a calculus course, let's use standard max/min procedures.
Think of a "general" (unspecified) point $P$ on the line $y=2x+3$. Suppose that the $x$-coordinate of $P$ is $x$. Then the $y$-coordinate of $P$ is $2x+3$. So $P$ is the point $(x,2x+3)$.
What is the distance from $(x,2x+3)$ to the origin? By what I hope is a familiar formula, this distance is $$\sqrt{(x-0)^2 +(2x+3-0)^2}.$$ This is because in general the distance between $(a,b)$ and $(c,d)$ is $\sqrt{(a-c)^2+(b-d)^2}$. Apply this formula, using $a=x$, $b=2x+3$, and $c=0$, $d=0$.
Alternately, you can find the distance from $(x,2x+3)$ to the origin by drawing a picture and using the Pythagorean Theorem.
We want to choose $x$ so as to minimize the distance from $P$ to the origin, so we want to minimize $\sqrt{x^2+(2x+3)^2}$.
We can now let $$f(x)=\sqrt{x^2+(2x+3)^2}$$ and find the $x$ that makes $f(x)$ smallest by the usual derivative process.
But there is a little trick that simplifies things. If $\sqrt{x^2+(2x+3)^2}$ is as small as possible, then so is $x^2+(2x+3)^2$, and vice-versa. So we try to minimize the square of the distance. Let $$g(x)=x^2+(2x+3)^2.$$ Can you now find the value of $x$ that makes $g(x)$ as small as possible?
Added: After some simplification, $$f'(x)= \frac{5x+6}{\left(x^2+(2x+3)^2\right)^{1/2}}.$$ If you take the suggested alternate route, you will find that $$g'(x)=10x+12.$$