# Homework: Stuck on the last step of find the critical values

Calculus section 4.4 Optimization problems

-Question: What two positive real number whose product is 19 have the smallest possible sum?

Let S be the sum of the two numbers. what is the objective functions in terms of one number, x?

-My answers so far are:

S= "$x+\frac{19}{x}$"

The interval of interest of the function is "(0, ∞)"

The numbers are __ and $\sqrt{19}$. (type the answers, using radicals as needed)

-My work: 19=x*y

$\frac{19}{x}=y$

S=x+y

$S=x+\frac{19}{x}$

$S'=1+\frac{-19}{x^2}$

$0=1+\frac{-19}{x^2}$

$-1=\frac{-19}{x^2}$

$1=\frac{19}{x^2}$

$1=19*\frac{1}{x^2}$

$\frac{1}{19}=\frac{1}{x^2}$ This is where I'm stuck.

Maybe I had to use $\frac{-b\pm\sqrt{b^2-4*a*c}}{a^2}$ to get a radical answer, but I can't tell where to find a, b, or c if I need to go that route.

Thank you all for your feedback, but then that still leaves the question what are the 2 numbers? I found 1 of them, $\sqrt{19}$ but other is not 0, or -$\sqrt{19}$.

• You don't need the quadratic formula. What you end up with is correct: just solve for $x$ by square rooting both sides. The result is $x=\sqrt{19}$. Commented Mar 27, 2016 at 19:36

That looks right to me, all you have to do is solve further $$\frac{1}{19}=\frac{1}{x^2}$$ $$19=x^2$$ $$x=\sqrt{19}$$ $$y=19/x=\sqrt{19}$$
• going from $\frac{1}{19}=\frac{1}{x^2}$ to 19 = $x^2$ is something i'm not fairly familiar with, would you mind explaining what happens right then? Commented Mar 27, 2016 at 19:38
• When solving equations you always apply the same operation to both sides. In going from $x+a=b$ to $x=b-a$ we subtracted $a$ from both sides. This holds for all operations: as long as you do the same thing on both sides, your equality remains valid. In particular, you can invert both sides: $a=b$ gives $\frac{1}{a}=\frac{1}{b}$ at least when $a$ and $b$ are not 0. Commented Mar 27, 2016 at 19:43
The numbers are $\sqrt{19}$ and $\sqrt{19}$ because S= $\sqrt{19}$ + $\frac{19}{\sqrt{19}}$
S= $\sqrt{19}$ + $\frac{19*\sqrt{19}}{19}$
S= $\sqrt{19}+\sqrt{19}$