# Minima maxima: a²y = x⁴

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## Yung Ans. Niya po is (0,0) at

Yung Ans. Niya po is (0,0) at Minimum.

Yung process sana po sir. Thanks po

## Yung process po sana kung

Yung process po sana kung paano solve.

Yung ans nya po is (0,0) , minimum.

## $a^2 y = x^4$

$a^2 y = x^4$

Differentiate

$a^2 y' = 4x^3$

Equate

y'= 0 to determine the critical points (maxima or minima)$a^2 (0) = 4x^3$

$x = 0$

For

x= 0$a^2 y = 0^4$

$y = 0$

Hence,

critical point = (0, 0)

Check the neighboring points to determine whether (0, 0) is minimum or maximum. Set

x= ±1:$a^2 y = (\pm 1)^4$

$y = +\dfrac{1}{a^2}$ ← above (0, 0)

Hence, the point (0, 0) is minimum.

## Salamat po sir

Salamat po sir

How about 9a²y=x(4a+x)³

Yung ans nya po is (a,3a) maximum

Thanks in advance sir

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