Weird Answer involving Minimum

Question

Given the isosceles triangle $ABC$, in which $AB$ is 12 inches and the altitude $CD$ is 3 inches, find the point $P$ on CD such that the sum of the distances of P from the vertices is a minimum.

This problem seemed fairly simple. Here's the equation that I set up:

$$2\sqrt{x^2+36} + 3 - x = d$$

Then, I took the derivative, set it to zero, and hoped to find the minimum: $$2 \cdot 2x \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{x^2+36}} -1 = d'$$ $$\frac{2x}{\sqrt{x^2+36}} = 1$$ $$4x^2 = x^2 + 36$$ $$x = 2\sqrt{3}$$

Firstly, my book says the answer should be where the point C lies (i.e, the length $x = 3$ inches). Secondly, $2\sqrt{3} > 3$.

If I'm right, I would like to know how to interpret this result (as I had obtained similar answers for previous questions). If I'm not right, I would like to know how I screwed up something so basic so I can then proceed to bash my head on the wall for the next few hours (not really).

Thanks.

Edit: Here's a better picture.

Answer 1

You do not explain how you derive your governing equation; things might have gone wrong somewhere there. Consider this:

If we make the point $A$ our origin of coordinates, then $B$ becomes $(12,0)$ and $C$ becomes $(m,3)$, where $0<m<12,$ indicates the length of $AD.$ The distances from our point $(m,y)$ from the three vertices is as follows: $$\begin{align}{|PA|=\sqrt{y^2+m^2}\\ |PB|=\sqrt{y^2+(m-12)^2}\\ |PC|=\sqrt{(y-3)^2}\\ }\end{align},$$ so that it is their sum that you want to minimize as $y$ varies in the bounds $[0,3].$

You should be able to continue from here?

Edit: When I took note of the fact that $|AC|=|CB|,$ so that consequently, $m=6,$ and simplified, differentiated and set to nullity, I got $$4y^2(y-3)^2=(3-y)^2(y^2+36).$$ It is now easy to see where OP lost his solution, since it is very tempting to cancel the equal factors; but recall that $3$ is in the domain of $y,$ so that we cannot legitimately cancel. Thus, letting $K=(3-y)^2,$ we get $$(4y^2-y^2-36)K=0,$$ from which we also get the possibility $y=3.$

Hope OP doesn't bang his head on a wall. :)

Answer 2

First observe that your function $d$ is a decreasing function of $x (\because$ its derivative is negative $\forall x\in[0,3])$.

Now, since we are restricted in this interval only, the required minimum occurs at $x=3$.

Hope it helps:)

Answer 3

Alright, so I think I understand why the answer is $x = 3$. If my equation is correct, then what I've found is the relative minimum of the function. However, we must find the absolute minimum, i.e, the minimum of the function in between the interval [a, b] (in this case $a=0$ and $b=3$). Therefore, because $b < 2\sqrt{3}$, the solution within the interval [0, 3] is $x = 3$. This point was made by @Allawonder and @Martund

If my reasoning is incorrect, then I would love feedback.