# Minima maxima: a²y = x⁴

How to solve 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

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

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