Question:
Find the max or min value of the function
$$y = x + \frac{16}{x+3}$$
where $x > 0$, without using differentiation.
I don't know how to solve this. How do I determine if it is a minimum or maximum?
Asked
Active
Viewed 93 times
-1
-
1You can have a look at How to prove this inequality $ x + \frac{1}{x} \geq 2 $ and other related questions. – Martin Sleziak Jun 15 '15 at 11:26
-
Thanks a lot. Martin. I can understand how to prove x+1/x >= 2 now – alanacv Jun 15 '15 at 14:48
-
The body of your Question should contain a full statement of the problem, not relying wholly on the title to pose it. See [ask]. – hardmath Jun 15 '15 at 21:39
4 Answers
4
Note that there is no max value for $f(x)$, but a min value does exist. Observe that for $x > 0$, $$ f(x) = x + \frac{16}{x+3} = \color{red}{x + 3 + \frac{16}{x+3}} - 3 \geq \color{red}{8} - 3 = 5 $$ When $x+3 = 4$, the equality holds, thus the min value is $5$.
Here, the fact $a + b \geq 2\sqrt{ab}, a, b > 0$ is used.
PSPACEhard
- 10,283
4
Your expression is non-negative for $x>0.$
First write the function as $$y=\frac{x^2+3x+16}{x+3},$$ and keep subtracting and adding $x+3$ from the numerator until it is factorable. This gives $$y=\frac{x^2-2x+1}{x+3}+5.$$ The zero of the fraction is $x=1,$ giving minimum value $y=5.$
John Molokach
- 1,885
- 15
- 19
-
Nice +1. You may want to write that as $\dfrac{(x-1)^2}{x+3}+5$, so it self-certifies the minimum for $\mathbb R^+$. – Macavity Jun 15 '15 at 11:47
2
\begin{align} x &> 0 \\ x + 3 &> 3 \\ y &= x + \frac{16}{x + 3} \end{align}
$$ \text{A little trick:} $$
$$ y + 3 = x + 3 + \frac{16}{x + 3} $$
$$ \text{It is easy to prove that } $$
\begin{align} a, b &> 0 \\ a + b &\geq 2\sqrt{ab} \end{align}
$$ \text {so now:} $$
\begin{align} a + b &\geq 2\sqrt{ab} \\ (x + 3) + \frac{16}{x + 3} &\geq 2\sqrt{(x + 3) \times \frac{16}{x + 3}} = 2 \times 4 = 8 \\ y + 3 = (x + 3) + \frac{16}{x + 3} &\geq 8 \\ y + 3 &\geq 8 \\ y &\geq 5 \end{align}
Aryan
- 442
Khosrotash
- 24,922
-
1Maybe it is important to add that $y=5$ is indeed attained (for $x+3=4$, i.e., for $x=1$; this fulfills the requirement $x>0$.) – Martin Sleziak Jun 15 '15 at 11:28
-
-
1
$$y=x+\frac{16}{x+3}\\(x+3)y=x(x+3)+16\\x^2+3x+16-xy-3y=0\\x^2+x(3-y)+(16-3y)=0\\ \Delta \geq0\\(3-y)^2-4(1)(16-3y) \geq 0\\y^2+9-6y-64+12y \geq0\\(y-5)(y+11) \geq 0\\y \leq-11 \space\space or \space \space y\geq 5\\ $$ as we know about $x> 0 \rightarrow y >0$ so $$ y\geq 5 $$
Khosrotash
- 24,922