Maximum value

Level 2

The maximum value of the expression $2\sin x + 4\cos x + 3$ can be expressed as $a\sqrt{b} + c$ , where $a$ , $b$ and $c$ are natural numbers. Find $a+b+c$ .

The answer is 10.

1 solution

Kenny Lau
Jan 3, 2014

The maximum of $2\sin x+4\cos x+3$ occurs whenever $\frac d{dx}(2\sin x+4\cos x+3)=0$ . $\begin{array}{rcl} \frac d{dx}(2\sin x+4\cos x+3)&=&0\\ 2\cos x-4\sin x&=&0\\ \cos x&=&2\sin x\\ \tan x&=&0.5 \end{array}$ In this photo , we can see that whenever $\tan x=0.5$ , $\sin x=\frac1{\sqrt5}$ and $\cos x=\frac2{\sqrt5}$ .

Therefore, $2\sin x+4\cos x+3=\frac2{\sqrt5}+\frac8{\sqrt5}+3=\frac{10}{\sqrt5}+3=2\sqrt5+3$

The solution is therefore 2+5+3 which is equal to $\boxed{10}$ .

Le maximum de $2\sin x+4\cos x+3$ se produit quand $\frac d{dx}(2\sin x+4\cos x+3)=0$ . $\begin{array}{rcl} \frac d{dx}(2\sin x+4\cos x+3)&=&0\\ 2\cos x-4\sin x&=&0\\ \cos x&=&2\sin x\\ \tan x&=&0.5 \end{array}$ Dans cette photo , on peut voir que quand $\tan x=0.5$ , $\sin x=\frac1{\sqrt5}$ et $\cos x=\frac2{\sqrt5}$ .

$2\sin x+4\cos x+3$ est donc $\frac2{\sqrt5}+\frac8{\sqrt5}+3=\frac{10}{\sqrt5}+3=2\sqrt5+3$

La réponse est donc 2+5+3 qui est égale à $\boxed{10}$ .

You could also solve for $x$ from $\cos x = 2 \sin x$ with the identity $\sin^2 x + \cos^2 x = 1$ .

Josh Petrin - 7 years, 5 months ago

Thanks :)

Kenny Lau - 7 years, 5 months ago

Yes, that would be easier. :)

Akshat Jain - 7 years, 5 months ago

More simply, from Cauchy Schwartz inequality, $\begin{array}{lcl} 2 \sin x + 4 \cos x &\leq& \sqrt{\left ( 2^2+4^2 \right ) \left ( \sin^2 x + \cos^2 x \right )} \\ &= & 2 \sqrt{5} \end{array}$ Equality holds when $\frac{ \sin x}{\cos x}= \frac{2}{4} = \frac{1}{2}$ $\implies x= \tan^{-1} \left ( \frac{1}{2} \right )$

Sreejato Bhattacharya - 7 years, 5 months ago

Nice idea of differentiation, Kenny. :)

Akshat Jain - 7 years, 5 months ago

Maximum value

The answer is 10.

1 solution

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