What is the extreme f(x)=(x/5)^1*(y/5)^{x-1} ?

The extreme f(x)=(x/5)^1*(y/5)^{x-1} is Saddle(1, 5/e)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=(x/5)^1*(y/5)^{x-1}

extreme points

f (x) = (\frac{x}{5})^{1} \cdot (\frac{y}{5})^{x - 1}

Saddle (1, \frac{5}{e})

Solution steps

Find the critical points:

(1, \frac{5}{e})

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{x y ^{x - 3} ( x - 1 ) ( x - 2 )}{5 ^{x}}

D (x, y) = \frac{x ^{4} y ^{2 x - 4} ln ( y ) ln ( \frac{y}{5} ) - ln ( 5 ) x ^{4} y ^{2 x - 4} ln ( \frac{y}{5} ) - x ^{4} y ^{2 x - 4} ln ^{2} ( \frac{y}{5} ) - 3 x ^{3} y ^{2 x - 4} ln ( \frac{y}{5} ) + x ^{3} y ^{2 x - 4} ln ( y ) - 3 x ^{3} y ^{2 x - 4} ln ( y ) ln ( \frac{y}{5} ) + 3 ln ( 5 ) x ^{3} y ^{2 x - 4} ln ( \frac{y}{5} ) - ln ( 5 ) x ^{3} y ^{2 x - 4} + 2 x ^{3} y ^{2 x - 4} ln ^{2} ( \frac{y}{5} ) + 3 x ^{2} y ^{2 x - 4} ln ( \frac{y}{5} ) - 3 x ^{2} y ^{2 x - 4} ln ( y ) + 2 x ^{2} y ^{2 x - 4} ln ( y ) ln ( \frac{y}{5} ) - 2 ln ( 5 ) x ^{2} y ^{2 x - 4} ln ( \frac{y}{5} ) + 3 ln ( 5 ) x ^{2} y ^{2 x - 4} - x ^{2} y ^{2 x - 4} ln ^{2} ( \frac{y}{5} ) - 4 x ^{2} y ^{2 x - 4} + 4 x y ^{2 x - 4} ln ( y ) - 4 ln ( 5 ) x y ^{2 x - 4} - 2 x y ^{2 x - 4} ln ( \frac{y}{5} ) + 4 x y ^{2 x - 4} - y ^{2 x - 4}}{2 5 ^{x}}

Check the sign of at each critical point

Check critical point

(1, \frac{5}{e}) :

Saddle

Saddle (1, \frac{5}{e})

f (x, y) = x^{2} + x y + y^{2} + 3 x - 3 y + 4

f (x) = x^{3} - 3 x^{2} + 10

f (x) = (x^{2} - 3) e^{x}

y = x^{4} - 4 x^{3} + 2

\frac{x + 3}{x ^{2}}

What is the extreme f(x)=(x/5)^1*(y/5)^{x-1} ?
The extreme f(x)=(x/5)^1*(y/5)^{x-1} is Saddle(1, 5/e)