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Question Number 78425 by arkanmath7@gmail.com last updated on 17/Jan/20

h o w t o s o l v e

f i n d i n f a n d s u p o f A

1 . A = {\frac{m}{n} + \frac{4 n}{m} m, n \in N}

2 . A = {\frac{m n}{4 m^{2} + n^{2}} m \in Z, n \in N}

3 . A = {\frac{m}{∣ m ∣ + n} m \in Z, n \in N}

Answered by MJS last updated on 17/Jan/20

\frac{d}{d m} [\frac{m}{n} + \frac{4 n}{m}] = \frac{1}{n} - \frac{4 n}{m^{2}} = 0 \Rightarrow m = \pm 2 n

\frac{d^{2}}{{d m}^{2}} [\frac{m}{n} + \frac{4 n}{m}] = \frac{8 n}{m^{3}} = {\begin{cases} - \frac{1}{n^{2}}; m = - 2 n \\ \frac{1}{n^{2}}; m = + 2 n \end{cases}

local \max at m = - 2 n but m, n \in N \Rightarrow

\Rightarrow A_{\max} = + \infty [m = 1 \land n \to + \infty or n = 1 \land m \to + \infty \Rightarrow A \to + \infty]

local \min at m = 2 n \land m, n \in N \Rightarrow A_{\min} = 4

Commented by MJS last updated on 17/Jan/20

because at a \max the 2^{nd} derivate is smaller

than 0

Commented by arkanmath7@gmail.com last updated on 17/Jan/20

w h y d i d y o u t a k e l o c a l m a x a t

m = - 2 n n o t a t m = 2 n ? ?

Commented by arkanmath7@gmail.com last updated on 17/Jan/20

t h a n k y o u s o m u c h s i r

Answered by MJS last updated on 17/Jan/20

\frac{d}{d m} [\frac{m n}{4 m^{2} + n^{2}}] = - \frac{n (4 m^{2} - n^{2})}{{(4 m^{2} + n^{2})}^{2}} = 0 \Rightarrow m = \pm \frac{n}{2}

\frac{d^{2}}{{d m}^{2}} [\frac{m n}{4 m^{2} + n^{2}}] = \frac{8 m n (4 m^{2} - 3 n^{2})}{{(4 m^{2} + n^{2})}^{3}} = {\begin{cases} \frac{1}{n^{2}}; m = - \frac{n}{2} \\ - \frac{1}{n^{2}}; m = \frac{n}{2} \end{cases}

\min at m = - \frac{n}{2}; A = - \frac{1}{4}

\max at m = \frac{n}{2}; A = \frac{1}{4}

try

\frac{m n}{4 m^{2} + n^{2}} < - \frac{1}{4}

4 m n < - 4 m^{2} - n^{2}

4 m^{2} + 4 m n + n^{2} < 0

{(2 m + n)}^{2} < 0 false

\frac{m n}{4 m^{2} + n^{2}} > \frac{1}{4} leads to {(2 m - n)}^{2} < 0 \Rightarrow false

Answered by MJS last updated on 17/Jan/20

- 1 < \frac{m}{∣ m ∣ + n} < + 1

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