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Question Number 204152 by universe last updated on 07/Feb/24

the maximum value of f (x, y) = xy - x^{3} y^{2}

attained over the square 0 ⩽ x ⩽ 1; 0 ⩽ y ⩽ 1 is

Answered by ajfour last updated on 08/Feb/24

f (x, y) = z = t - t^{3}

\frac{d z}{d t} = 1 - 3 t^{2} = 0 \Rightarrow t = \frac{1}{\sqrt{3}}

z_{m a x} = \frac{2}{\sqrt{27}} \approx 0.38490

Answered by mr W last updated on 08/Feb/24

f (1, y) = y - y^{2} = \frac{1}{4} - {(y - \frac{1}{2})}^{2}

f {(1, y)}_{m a x} = \frac{1}{4} = 0.25

f (x, 1) = x - x^{3}

\frac{d f (x, 1)}{d x} = 1 - 3 x^{2} = 0 \Rightarrow x = \frac{1}{\sqrt{3}}

f {(x, 1)}_{m a x} = \frac{1}{\sqrt{3}} - \frac{1}{3 \sqrt{3}} = \frac{2 \sqrt{3}}{9} \approx 0.3849

\Rightarrow f {(x, y)}_{m a x} = \frac{2 \sqrt{3}}{9} ✓

Commented by ajfour last updated on 08/Feb/24

t h i n k i m i s r e a d t h e q u e s t i o n p h o n e s c r e e n

w a s n o t c l e a n b u t t h i s g a v e y o u s o m e

c l u e!

Commented by mr W last updated on 08/Feb/24

h o w i c a m e t o t h e p a t h :

s i n c e t h e r e i s n o s o l u t i o n f o r

\frac{\partial f}{\partial x} = 0 \land \frac{\partial f}{\partial y} = 0 e x c e p t a t x = y = 0,

t h a t m e a n s t h e r e i s n o e x t r e m u m

a n d w e s h o u l d s e a r c h t h e

m a x i m u m o n t h e b o u n d a r y :

x = 0 o r x = 1 o r y = 0 o r y = 1 .

x = 0 o r y = 0 g i v e s f = 0

x = 1 g i v e s f_{m a x} = \frac{1}{4}

y = 1 g i v e s f_{m a x} = \frac{2 \sqrt{3}}{9}

\Rightarrow f_{m a x} = \frac{2 \sqrt{3}}{9} f o r 0 ⩽ x, y ⩽ 1

Commented by universe last updated on 08/Feb/24

t h a n k s s i r