What is the smallest possible value of their sum?

Question

The product of two positive numbers is 36. What is the smallest possible value of their sum?

so far I got

$$xy=36$$

$$y=\frac{36}{x}$$

Answer 1

Hint: $x+y \geq 2\sqrt{xy}$. Can you finish it?

Answer 2

OK, so you have the two numbers $x$ and $\frac{36}{x}$. You want to minimise their sum, so you want to minimise the expression $x + \frac{36}{x}$. Remember that to minimise an expression you take the derivative and set it equal to zero. Do you think you can solve it from there?

Answer 3

HINT: Note that $4xy=(x+y)^2-(x-y)^2$ or $(x+y)^2=4xy+(x-y)^2$

Answer 4

Hint: Consider the function $f(x)=x+\frac{36}{x}$ (i.e. the sum of $x$ and $y$) and find the minimum for that function.

Answer 5

Note: there are many ways you can show this but i am assuming from your tag you need an answer via derivatives here it goes

let $m=x+y=x+\frac{36}x=x+36x^{-1}$

then $m'=1-36x^{-2}$

to have a minimum value of $m$ let $m'=0$

$1-36x^{-2}=0$

$36=x^2$

so $x=6$ and $y=6$

Answer 6

xy=36, so 2xy=72. By $x^2-2xy+y^2 \ge 0$, $x^2+y^2 \ge 72$, so $x^2+2xy+y^2 \ge 72+36\cdot2=144$. Thus $x+y \ge 12$, since they are positive.

Answer 7

First note that $$f(x)=x+\frac{36}{x}=x+36x^{-1}$$ $$\frac{d}{dx}[f(x)]=1-36x^{-2}=1-\frac{36}{x^2}$$ So now let's find the critical points $$\frac{d}{dx}[f(x)]=0 $$ $$ 1-\frac{36}{x^2}=0 $$ $$ 1=\frac{36}{x^2} $$ $$ x^2=36= (\pm 6)^2 $$ $$ x=\sqrt{(\pm 6)^2}=\pm 6 $$ Since we're only considering positive values, we'll omit the negative value. Next we must check if $x=6$ is a local minima of $f(x)$. So now $$\frac{d^2}{dx^2}[f(x)]=72x^{-3}=\frac{72}{x^3}$$ $$\frac{d^2}{dx^2}[f(6)]=\frac{72}{6^3}\gt 0 $$ Therefore, $f(x)=x+\frac{36}{x}$ has a local minima at $x=6$. Which implies that the smallest sum of positive values is $$f(6)=6+6=12$$