Really simple

Algebra Level 4

Let a be the minimum value of $N$ and $b$ is the value of $x$ when the equality holds, given that $x>1$ $N=\frac{x+8}{\sqrt{x-1}}$

Find $a+b$ .

The answer is 16.

4 solutions

P C
Jan 13, 2016

Using AM-GM we get $x-1+9\geq2\sqrt{9(x-1)}$ $\Leftrightarrow x+8\geq6\sqrt{x-1}$ $\Leftrightarrow\frac{x+8}{\sqrt{x-1}}\geq6$ The equality holds when $x=10$ $\Rightarrow 10+6=16$

Can you explain why x=10

Son Nguyen - 5 years, 5 months ago

The equality holds when $x-1=9\Leftrightarrow x=9+1=10$

P C - 5 years, 5 months ago

How do you know that x=10.(I mean that you can predict x when the inequality holds.)

Son Nguyen - 5 years, 4 months ago

A Former Brilliant Member
Jan 15, 2016

First, we differentiate the equation. We get $\frac{dN}{dx}=\frac{x-10}{2(x-1)^{\frac{3}{2}}}=0$ by a combination of the chain rule and quotient rule.

Solving, $x=10$ and substituting it back into $N$ we get the required answer.

Nice solution

Jun Arro Estrella - 5 years, 3 months ago

@Jerry Han Jia Tao How do you know that it is the minimum value? The derivative of a function is zero even at its maximum.

Bug Menot - 5 years, 2 months ago

Of course one must check to see if the local extrema is the maximum or the minimum value.

A Former Brilliant Member - 5 years, 2 months ago

Edwin Gray
Feb 23, 2019

If N =(x + 8)/[(x - 1)^(1/2)], dN/dx = 0 gives x - 1 - x/2 - 4 = 0, or x/2 = 5, x = 10. Then N = 18/3 = 6, and 10 + 6 = 16.

Aaghaz Mahajan
Jul 8, 2018

Or, to simplify things even more................Use trigo subs.......!!!

Really simple

The answer is 16.

4 solutions

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